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# Solving Trigonometric Equations

This section covers:

Solving trig equations is just finding the solutions of equations like we did with linear, quadratic, and radical equations, but using trig functions instead. We will mainly use the Unit Circle to find the exact solutions if we can, and we’ll start out by finding the solutions from $$\left[ 0,2\pi \right)$$.

We can also solve these using a Graphing Calculator, as we’ll see below.

Note that we will use Trigonometric Identities to solve trig problems in the Trigonometric Identity section.

Important Note: There is a subtle distinction between finding inverse trig functions and solving for trig functions. If we want $$\displaystyle {{\sin }^{{-1}}}\left( {\frac{{\sqrt{2}}}{2}} \right)$$ for example, like in the The Inverse Trigonometric Functions section, we only pick the answers from Quadrants I and IV, so we get  $$\displaystyle \frac{\pi }{4}$$ only. But if we are solving $$\displaystyle \sin \left( x \right)=\frac{{\sqrt{2}}}{2}$$ we get $$\displaystyle \frac{\pi }{4}$$ and $$\displaystyle \frac{{3\pi }}{4}$$  in the interval $${\left[ {0,2\pi } \right)}$$; there are no domain restrictions.

# Solving Trigonometric Equations Using the Unit Circle

Let’s start out with solving fairly simple Trig Equations and getting the solutions from $$\left[ 0,2\pi \right)$$, or $$\left[ {0,{{{360}}^{o}}} \right)$$.

Here is the Unit Circle again so we can “pick off” the answers from it:

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Notice how sometimes we have to divide up the equation into two separate equations, like when the argument of the trig function is an expression, like $$\displaystyle \theta +\frac{\pi }{{18}}$$. Also note that $${{\left( \cos \theta \right)}^{2}}$$  is written as $${{\cos }^{2}}\theta$$, and we can put it in the graphing calculator as $$\boldsymbol{\cos {{\left( x \right)}^{2}}}$$ or $$\boldsymbol {{{\left( {\cos \left( x \right)} \right)}^{2}}}$$.

If the coefficient before the $$\theta$$ is less than 1, we may have to “throw away” an extraneous solution (like in the last tan problem below). If we have a coefficient before the $$\theta$$ that is greater than 1, we have to first find the general solution of the equation, and then go back to the Unit Circle to see where the solutions are in the $$\left[ 0,2\pi \right)$$ interval. We will learn how to do this here.

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Note that sometimes you may have to solve using degrees $$\left[ {0,{{{360}}^{o}}} \right)$$ instead of radians. Also note that sometimes we have to divide a sin by a cos to get a tan, as in one of the examples. And the last problem involves solving a trig inequality.

 Solving Trig Problems:  $$\boldsymbol {\left[ {0,2\pi } \right)}$$ $$\displaystyle \sin \theta =1$$ $$\theta =\left\{ {\frac{\pi }{2}} \right\}$$   $$\displaystyle \sin \theta =0$$ $$\theta =\left\{ {0,\pi } \right\}$$   $$\displaystyle \cos \theta =0$$ $$\theta =\left\{ {\frac{\pi }{2},\frac{{3\pi }}{2}} \right\}$$ $$\tan \theta +1=0$$ $$\displaystyle \begin{array}{c}\tan \theta =-1\\\theta =\left\{ {\frac{{3\pi }}{4},\frac{{7\pi }}{4}} \right\}\end{array}$$   $$\tan \theta -1=0$$ $$\displaystyle \begin{array}{c}\tan \theta =1\\\theta =\left\{ {\frac{\pi }{4},\frac{{5\pi }}{4}} \right\}\end{array}$$ $$\displaystyle \begin{array}{c}3\sqrt{2}\cos \theta +3=0\\\text{(degrees)}\end{array}$$   $$\displaystyle \begin{array}{c}3\sqrt{2}\cos \theta =-3\\\cos \theta =-\frac{3}{{3\sqrt{2}}}\\\cos \theta =-\frac{1}{{\sqrt{2}}}=-\frac{{\sqrt{2}}}{2}\\\theta =\left\{ {135{}^\circ ,225{}^\circ } \right\}\end{array}$$ $$2\sin \theta +4=3$$   $$\begin{array}{c}2\sin \theta =-1\\\sin \theta =-\frac{1}{2}\\\theta =\left\{ {\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right\}\end{array}$$ $$\displaystyle 4\sec \theta +7=-1$$   $$\displaystyle \begin{array}{c}4\sec \theta =-8\\\sec \theta =-2\,\,\,\,\,\left( {\cos \theta =-\frac{1}{2}} \right)\\\theta =\left\{ {\frac{{2\pi }}{3},\frac{{4\pi }}{3}} \right\}\end{array}$$ $$\displaystyle 2{{\cos }^{2}}\theta =1$$   $$\displaystyle \begin{array}{c}{{\cos }^{2}}\theta =\frac{1}{2}\\\sqrt{{{{{\cos }}^{2}}\theta }}=\sqrt{{\frac{1}{2}}}\\\cos \theta =\pm \frac{1}{{\sqrt{2}}}=\pm \frac{{\sqrt{2}}}{2}\\\theta =\left\{ {\frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4}} \right\}\end{array}$$ $$\displaystyle \sin \left( {\theta +\frac{\pi }{{18}}} \right)=0$$   $$\displaystyle \begin{array}{c}\theta +\frac{\pi }{{18}}=0\,\,\,\,\,\,\theta +\frac{\pi }{{18}}=\pi \\\,\,\,\theta =-\frac{\pi }{{18}}\,\,\,\,\,\,\,\theta =\frac{{17\pi }}{{18}}\\\text{Since}-\frac{\pi }{{18}}\text{ isn }\!\!’\!\!\text{ t in}\left[{0,\text{2}\pi } \right),\\\left\{ {-\frac{\pi }{{18}}+2\pi ,\frac{{17\pi }}{{18}}} \right\}\\=\left\{ {\frac{{17\pi }}{{18}},\frac{{35\pi }}{{18}}} \right\}\end{array}$$ $$\displaystyle \sin \left( {\frac{\theta }{3}+\frac{\pi }{3}} \right)=\cos \left( {\frac{\theta }{3}+\frac{\pi }{3}} \right)$$   $$\displaystyle \begin{array}{c}\tan \left( {\frac{\theta }{3}+\frac{\pi }{3}} \right)=1\\\frac{\theta }{3}+\frac{\pi }{3}=\frac{\pi }{4}\,\,\,\,\,\,\,\,\,\,\frac{\theta }{3}+\frac{\pi }{3}=\frac{{5\pi }}{4}\\\,\,\,\frac{\theta }{3}=-\frac{\pi }{{12}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{\theta }{3}=\frac{{11\pi }}{{12}}\\\,\,\,\theta =-\frac{\pi }{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta =\frac{{33\pi }}{{12}}\end{array}$$   Neither are in $$\left[ {0,2\pi } \right)$$, and adding $$2\pi$$ to $$\displaystyle -\frac{\pi }{4}$$ doesn’t work (extraneous); no solution, or $$\emptyset$$. Here’s one where we have an inequality:   For what values of $$\theta$$ is $$2\cos \theta <-\sqrt{3}$$ on the $$\left[ {0,2\pi } \right)$$ interval? $$\displaystyle 2\cos \theta <-\sqrt{3}$$ $$\displaystyle \cos \theta <-\frac{{\sqrt{3}}}{2}$$ This is a little tricky; if we look on the Unit Circle, we want the cos values to be less than where $$\displaystyle \cos \theta =-\frac{{\sqrt{3}}}{2}$$. Since when $$\displaystyle \cos \theta =-\frac{{\sqrt{3}}}{2},\,\,\theta =\left\{ {\frac{{5\pi }}{6},\frac{{7\pi }}{6}} \right\}$$, the $$x$$ will be less (or to the left of) of those 2 values, so we can see that $$\displaystyle \frac{{5\pi }}{6}<\theta <\frac{{7\pi }}{6}\text{ or }\left( {\frac{{5\pi }}{6},\frac{{7\pi }}{6}} \right)$$.   You can also plug in values for cos on the Unit Circle to see which are less than $$\displaystyle -\frac{{\sqrt{3}}}{2}$$, or use a Graphing Calculator to graph the two sides of the inequality.

# Solving Trigonometric Equations – General Solutions

Since trig functions go on and on in both directions of the $$x$$-axis, we’ll also have to know how to solve trig equations over the set of real numbers; this is called finding the general solutions for these equations.

We still use the Unit Circle to do this, but we have to think about adding and subtracting multiples of $$2\pi$$ for the sin, cos, csc, and sec functions (since $$2\pi$$ is the period for them), and $$\pi$$ for the tan and cot functions (since $$\pi$$ is the period for them). We can do this by adding $$2\pi k$$ or $$\pi k$$ where $$k$$ is any integer (positive or negative).

Also note that a lot of times, when we get the solutions for tan, they are 180° or $$\pi$$ radians apart, so one set of solutions will the same as the other, and we can collapse into one solution and add $$\pi k$$. This will sometimes happen if trig functions are squared in the problems also, since we’ll getting plusses and minuses.

Here are examples; find the general solution, or all real solutions) for the following equations. Note that $$k$$ represents all integers $$\left( k\in \mathbb{Z} \right)$$. Note also that I’m using “fancy” notation; you may not be required to do this.

Also note that sometimes you’ll solve for $$x$$ or another variable, sometimes for $$\theta$$, depending on your book or teacher.

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Note that we need to be careful about domain restrictions with our answers. For tan, cot, csc, and sec, we have asymptotes, and if our answer happens to fall on an asymptote, we have to eliminate it.

 Solving Trig Problems:  General Solutions $$\displaystyle 6\cos \theta =3\sqrt{3}$$   $$\displaystyle \begin{array}{c}\cos \theta =\frac{{\sqrt{3}}}{2}\\\\\left\{ {\theta |\theta =\frac{\pi }{6}+2\pi k,\,\,\theta =\frac{{11\pi }}{6}+2\pi k} \right\}\end{array}$$ $$\displaystyle \begin{array}{c}\color{#800000}{{\sin x+3=4\sin x}}\\\\3\sin x=3\\\sin x=1\\\,\\\left\{ {x|x=\frac{\pi }{2}+\,\,2\pi k} \right\}\end{array}$$ $$\displaystyle \begin{array}{c}\color{#800000}{{\sqrt{3}\csc x=-2}}\\\text{csc}\,x=-\frac{2}{{\sqrt{3}}}\,\,\,\,\,\,\,\,\,\left( {\sin x=-\frac{{\sqrt{3}}}{2}} \right)\\\\\left\{ {x|x=\frac{{4\pi }}{3}+2\pi k,\,x=\frac{{5\pi }}{3}+2\pi k} \right\}\end{array}$$ $$\displaystyle 2\sec \left( {x-\frac{\pi }{4}} \right)=4$$   $$\displaystyle \begin{array}{c}\sec \left( {x-\frac{\pi }{4}} \right)=2\text{ }\left( {\cos \left( {x-\frac{\pi }{4}} \right)=\frac{1}{2}} \right)\\\\x-\frac{\pi }{4}=\frac{\pi }{3}+2\pi k;\,\,x-\frac{\pi }{4}=\frac{{5\pi }}{3}+2\pi k\\\left\{ {x|x=\frac{{7\pi }}{{12}}+2\pi k,\,\,x=\frac{{23\pi }}{{12}}+2\pi k} \right\}\end{array}$$ $$\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}4\sec \theta +10=-\sec \theta \\\text{(degrees)}\end{array}}\\\\5\sec \theta =-10\\\sec \theta =-2\text{ }\left( {\cos \theta =-\frac{1}{2}} \right)\\\\\left\{ \begin{array}{l}\theta |\theta =120{}^\circ +360{}^\circ k,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,240{}^\circ +360{}^\circ k\end{array} \right\}\end{array}$$ $$\displaystyle \begin{array}{c}\color{#800000}{{2\sin x=4\sin x}}\\\\2\sin x=0\\\sin x=0\\\\\{x|x=2\pi k,\,\,\,x=\pi +2\pi k\}\end{array}$$ Note that (by looking at Unit Circle) this can be simplified to $$\displaystyle \{x|x=\pi k\}$$ $$\displaystyle \text{co}{{\text{s}}^{2}}\theta =\frac{1}{2}$$   $$\displaystyle \begin{array}{c}\sqrt{{\text{co}{{\text{s}}^{2}}\theta }}=\sqrt{{\frac{1}{2}}}\\\cos \theta =\,\,\pm \frac{1}{{\sqrt{2}}}=\,\,\pm \frac{{\sqrt{2}}}{2}\\\\\{\theta |\theta =\frac{\pi }{4}+2\pi k,\,\,\,\theta =\frac{{3\pi }}{4}+2\pi k,\\\,\,\,\,\,\,\,\,\,\theta =\frac{{5\pi }}{4}+2\pi k,\,\,\theta =\frac{{7\pi }}{4}+2\pi k\}\end{array}$$ Note that (by looking at Unit Circle) this can be simplified to $$\{\theta |\theta =\frac{\pi }{4}+\pi k,\,\,\,\theta =\frac{{3\pi }}{4}+\pi k\}$$ $$\displaystyle \tan \left( {x+\frac{\pi }{2}} \right)=\sqrt{3}$$   $$\displaystyle \begin{array}{c}x+\frac{\pi }{2}=\frac{\pi }{3}\,\,\,\,\,\,\,\,\,\,\,\,x+\frac{\pi }{2}=\frac{{4\pi }}{3}\\\,\,\,x=-\frac{\pi }{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{5\pi }}{6}\end{array}$$ Note that (by looking at Unit Circle) this can be simplified to $$\displaystyle \{x|x=\frac{{5\pi }}{6}+\pi k\}$$   (Remember that you add $$\pi k$$ instead of $$2\pi k$$ for tan and cot). $$\displaystyle \begin{array}{c}\color{#800000}{{3{{{\cot }}^{2}}\theta =1}}\\\text{co}{{\text{t}}^{2}}\theta =\frac{1}{3}\\\sqrt{{{{{\cot }}^{2}}\theta }}=\sqrt{{\frac{1}{3}}}\\\cot \theta =\,\,\pm \frac{1}{{\sqrt{3}}}\\\\\{\theta |\theta =\frac{\pi }{3}+\pi k,\,\,\,\theta =\frac{{2\pi }}{3}+\pi k,\\\,\,\,\,\,\,\,\,\,\theta =\frac{{4\pi }}{3}+\pi k,\,\,\theta =\frac{{5\pi }}{3}+\pi k\}\end{array}$$ Note that (by looking at Unit Circle) this can be simplified to $$\{\theta |\theta =\frac{\pi }{3}+\pi k,\,\,\,\theta =\frac{{2\pi }}{3}+\pi k\}$$ $$\displaystyle \frac{{\cot \theta }}{4}=\sin \theta \cos \theta$$    $$\displaystyle \begin{array}{c}\frac{{\cos \theta }}{{4\sin \theta }}=\sin \theta \cos \theta \\\cos \theta =4{{\sin }^{2}}\theta \cos \theta \\4{{\sin }^{2}}\theta \cos \theta -\cos \theta =0\\\cos \theta \left( {4{{{\sin }}^{2}}-1} \right)=0\\\cos \theta \left( {2\sin -1} \right)\left( {2\sin \theta +1} \right)=0\\\cos \theta =0\text{ }\,\,\text{ }\,\,\sin \theta =\pm \frac{1}{2}\,\\\\\{\theta |\theta =\frac{\pi }{2}+\pi k,\,\,\,\theta =\frac{\pi }{6}+\pi k,\,\,\,\theta =\frac{{5\pi }}{6}+\pi k\,\}\end{array}$$ Watch domain restriction: $$\displaystyle \frac{{1-\cos x}}{{\sin x}}=\frac{{\sin x}}{{1+\cos x}}$$   $$\begin{array}{c}\left( {1-\cos x} \right)\left( {1+\cos x} \right)={{\sin }^{2}}x\\1-{{\cos }^{2}}x={{\sin }^{2}}x\\{{\sin }^{2}}x={{\sin }^{2}}x;\,\,\,\,\mathbb{R}\end{array}$$ We get all real numbers since this is an identity. But we still need to check the domain restrictions (denominator can’t be 0): $$\begin{array}{c}\sin x\ne 0,\,\,\,1+\cos x\ne 0\\\sin x\ne 0,\,\,\,\cos x\ne -1\\\text{Solution: }\left\{ {x|\,\,x\ne \pi +2\pi k} \right\}\end{array}$$

Note that you can check these in a graphing calculator (radian mode) by putting the left-hand side of the equation into $${{Y}_{1}}$$ and the right-hand side into $${{Y}_{2}}$$ and get the intersection. You won’t get the exact answers, but you can still compare to the exact answers you got above.

# Solving Trigonometric Equations with Multiple Angles

We have to be careful when solving trig equations with multiple angles, meaning there is a coefficient before the $$x$$ or $$\theta$$ (variable). This is because we could have fewer or more solutions in the Unit Circle, and thus for all real solutions when we add the $$2\pi k$$ or $$\pi k$$.

So when we solve these types of trig problems, we always want to solve for the General Solution first (even if we’re asked to get the solutions between 0 and $$2\pi k$$) and then go back and see how many solutions are on the Unit Circle (between 0 and $$2\pi k$$ ).

When solving trig equations with multiple angles between 0 and $$2\pi$$, we’ll typically get fewer solutions if the coefficient of the variable is less than 1, or more solutions if the coefficient of the variable is greater than 1. As an example, we typically get 2 solutions for $$\cos \left( \theta \right)$$  between 0 and $$2\pi$$, so for  $$\cos \left( 3\theta \right)$$, we’ll get 2 times 3, or 6 solutions. As another example, for $$\displaystyle \cos \left( \frac{\theta }{2} \right)$$, we’ll only get one solution instead of the normal two.

Let’s do some problems, finding the general solutions first, and then finding the solutions in the 0 to $$2\pi$$ interval.

Note that when we multiply or divide to get the variable by itself, we have to do the same with the “$$+2\pi k$$” or “$$+\pi k$$”.

Again, watch for domain restrictions; answers that happen to fall on an asymptote for tan, cot, sec, or csc.

 Solving Trig Problems with Multiple Angles – General Solutions $$\displaystyle \tan \left( {3\theta } \right)=\sqrt{3}$$   $$3\theta =\frac{\pi }{3}+\pi k\,\,\,\,\,\,\,\,3\theta =\frac{{4\pi }}{3}+\pi k$$   Looking at Unit Circle, this can be simplified to: $$\displaystyle \begin{array}{c}3\theta =\frac{\pi }{3}+\pi k\\\left\{ {\theta |\theta =\frac{\pi }{9}+\frac{\pi }{3}k} \right\}\end{array}$$ $$\displaystyle 2\cos \left( {3x} \right)+\sqrt{3}=0$$   $$\displaystyle \begin{array}{c}\cos \left( {3x} \right)=-\frac{{\sqrt{3}}}{2}\\\,\,\,3x=\frac{{5\pi }}{6}+2\pi k\,\,\,\,\,\,\,\,\,\,3x=\frac{{7\pi }}{6}+2\pi k\\\\\left\{ {x|x=\frac{{5\pi }}{{18}}+\frac{2}{3}\pi k,\,\,x=\frac{{7\pi }}{{18}}+\frac{2}{3}\pi k\,\,} \right\}\end{array}$$ $$\displaystyle \sqrt{2}\sec \left( {\frac{x}{6}} \right)-2=0$$   $$\displaystyle \begin{array}{c}\sec \left( {\frac{x}{6}} \right)=\frac{2}{{\sqrt{2}}}\,\,\,\,\,\,\,\,\left( {\cos \left( {\frac{x}{6}} \right)=\frac{{\sqrt{2}}}{2}} \right)\\\,\,\frac{x}{6}=\frac{\pi }{4}+2\pi k\,\,\,\,\,\,\,\,\,\,\,\,\frac{x}{6}=\frac{{7\pi }}{4}+2\pi k\\\,\,\,x=\frac{{6\pi }}{4}+12\pi k\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{42\pi }}{4}+12\pi k\\\\\left\{ {x|x=\frac{{3\pi }}{2}+12\pi k,\,\,\,\,x=\frac{{21\pi }}{2}+12\pi k\,\,} \right\}\end{array}$$ $$\displaystyle 5{{\cos }^{2}}\left( {\frac{\theta }{3}} \right)=5$$   $$\displaystyle \begin{array}{c}\sqrt{{{{{\cos }}^{2}}\left( {\frac{\theta }{3}} \right)}}=\sqrt{1}\\cos\left( {\frac{\theta }{3}} \right)=\pm 1\\\,\,\,\frac{\theta }{3}=0+2\pi k\,\,\,\,\,\,\,\,\frac{\theta }{3}=\pi +2\pi k\end{array}$$   Looking at Unit Circle, this can be simplified to: $$\displaystyle \begin{array}{c}\frac{\theta }{3}=\pi k\\\left\{ {\theta |\theta =3\pi k} \right\}\end{array}$$ $$\displaystyle 2{{\sin }^{2}}\left( {2x} \right)=1$$   $$\displaystyle \begin{array}{c}{{\sin }^{2}}\left( {2x} \right)=\frac{1}{2}\\\sqrt{{{{{\sin }}^{2}}\left( {2x} \right)}}=\sqrt{{\frac{1}{2}}}\\\sin \left( {2x} \right)=\,\,\pm \frac{1}{{\sqrt{2}}}=\,\,\pm \frac{{\sqrt{2}}}{2}\\\,\,\,2x=\frac{\pi }{4}+2\pi k\,\,\,\,\,\,\,\,\,\,\,\,2x=\frac{{3\pi }}{4}+2\pi k\\\,\,\,2x=\frac{{5\pi }}{4}+2\pi k\,\,\,\,\,\,\,\,\,\,2x=\frac{{7\pi }}{4}+2\pi k\end{array}$$   Looking at Unit Circle, this can be simplified to: $$\displaystyle \begin{array}{l}2x=\frac{\pi }{4}+\pi k\,\,\,\,\,\,\,\,\,\,\,\,2x=\frac{{3\pi }}{4}+\pi k\\\left\{ {x|x=\frac{\pi }{8}+\frac{\pi }{2}k,\,\,\,x=\frac{{3\pi }}{8}+\frac{\pi }{2}k} \right\}\end{array}$$ $$\displaystyle \tan \left( {\frac{\theta }{2}-\frac{\pi }{3}} \right)=-1$$   $$\frac{\theta }{2}-\frac{\pi }{3}=\frac{{3\pi }}{4}+\pi k\,\,\,\,\,\,\,\,\,\frac{\theta }{2}-\frac{\pi }{3}=\frac{{7\pi }}{4}+\pi k$$   Looking at Unit Circle, this can be simplified to: $$\displaystyle \begin{array}{c}\frac{\theta }{2}-\frac{\pi }{3}=\frac{{3\pi }}{4}+\pi k\\\\\frac{\theta }{2}=\left( {\frac{{3\pi }}{4}+\frac{\pi }{3}} \right)+\pi k\\\frac{\theta }{2}=\frac{{13\pi }}{{12}}+\pi k\\\\\left\{ {\theta |\theta =\frac{{13\pi }}{6}+2\pi k} \right\}\,\text{or}\,\left\{ {\theta |\theta =\frac{\pi }{6}+2\pi k} \right\}\end{array}$$

Now let’s solve the same multiple angle problems, but get solutions between 0 and $$2\pi k$$.

Note again that when we solve these types of trig problems, we always want to solve for the General Solution first, and then go back and see how many solutions are on the Unit Circle (between 0 and $$2\pi k$$).

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 Solving Trig Problems with Multiple Angles:  $$\boldsymbol{\left[ {0,2\pi } \right]}$$ $$\displaystyle \tan \left( {3\theta } \right)=\sqrt{3}$$   $$3\theta =\frac{\pi }{3}+\pi k\,\,\,\,\,\,\,\,3\theta =\frac{{4\pi }}{3}+\pi k$$   Looking at Unit Circle, this can be simplified to: $$\displaystyle \begin{array}{c}3\theta =\frac{\pi }{3}+\pi k\\\,\theta =\frac{\pi }{9}+\frac{\pi }{3}k\,\,=\,\,\frac{\pi }{9}+\frac{{3\pi }}{9}k\end{array}$$   Looking at the Unit Circle, we can get solutions between 0 and $$2\pi$$ by adding $$\frac{{3\pi }}{9}$$, staying under $$2\pi$$: $$\theta =\left\{ {\frac{\pi }{9},\frac{{4\pi }}{9},\frac{{7\pi }}{9},\frac{{10\pi }}{9},\frac{{13\pi }}{9},\frac{{16\pi }}{9}} \right\}$$ $$\displaystyle 2\cos \left( {3x} \right)+\sqrt{3}=0$$   $$\displaystyle \begin{array}{c}\cos \left( {3x} \right)=-\frac{{\sqrt{3}}}{2}\\\,\,\,3x=\frac{{5\pi }}{6}+2\pi k\,\,\,\,\,\,\,\,\,3x=\frac{{7\pi }}{6}+2\pi k\\\,\,\,x=\frac{{5\pi }}{{18}}+\frac{{12}}{{18}}\pi k\,\,\,\,\,\,\,\,\,\,x=\frac{{7\pi }}{{18}}+\frac{{12}}{{18}}\pi k\end{array}$$   Looking at the Unit Circle, we can get solutions between 0 and $$2\pi$$ by adding $$\frac{{12\pi }}{{18}}$$, staying under $$2\pi$$: $$x=\left\{ {\frac{{5\pi }}{{18}},\frac{{17\pi }}{{18}},\frac{{29\pi }}{{18}},\frac{{7\pi }}{{18}},\frac{{19\pi }}{{18}},\frac{{31\pi }}{{18}}} \right\}$$ $$\displaystyle \sqrt{2}\sec \left( {\frac{x}{6}} \right)-2=0$$   $$\displaystyle \begin{array}{c}\sec \left( {\frac{x}{6}} \right)=\frac{2}{{\sqrt{2}}}\,\,\,\,\,\,\,\left( {\cos \left( {\frac{x}{6}} \right)=\frac{{\sqrt{2}}}{2}} \right)\\\,\,\frac{x}{6}=\frac{\pi }{4}+2\pi k\,\,\,\,\,\,\,\,\,\,\,\frac{x}{6}=\frac{{7\pi }}{4}+2\pi k\\\,\,\,x=\frac{{6\pi }}{4}+12\pi k\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{42\pi }}{4}+12\pi k\end{array}$$   Looking at the Unit Circle, the only solution that will work is $$\frac{{6\pi }}{4}=\frac{{3\pi }}{2}$$. $$x=\left\{ {\frac{{3\pi }}{2}} \right\}$$ $$\displaystyle 5{{\cos }^{2}}\left( {\frac{\theta }{3}} \right)=5$$   $$\displaystyle \begin{array}{c}\sqrt{{{{{\cos }}^{2}}\left( {\frac{\theta }{3}} \right)}}=\sqrt{1}\\cos\left( {\frac{\theta }{3}} \right)=\pm 1\\\,\,\,\frac{\theta }{3}=0+2\pi k\,\,\,\,\,\,\,\,\,\,\frac{\theta }{3}=\pi +2\pi k\end{array}$$   Looking at the Unit Circle, this can be simplified to: $$\displaystyle \begin{array}{c}\frac{\theta }{3}=\pi k\\\theta =3\pi k\end{array}$$   Looking at the Unit Circle, the only solution that will work is 0. $$\displaystyle \theta =\left\{ 0 \right\}$$ $$\displaystyle 2{{\sin }^{2}}\left( {2x} \right)=1$$   $$\displaystyle \begin{array}{c}\sqrt{{{{{\sin }}^{2}}\left( {2x} \right)}}=\sqrt{{\frac{1}{2}}}\\\sin \left( {2x} \right)=\,\,\pm \frac{1}{{\sqrt{2}}}=\,\,\pm \frac{{\sqrt{2}}}{2}\\\,\,\,2x=\frac{\pi }{4}+2\pi k\,\,\,\,\,\,\,\,\,\,2x=\frac{{3\pi }}{4}+2\pi k\\\,\,\,2x=\frac{{5\pi }}{4}+2\pi k\,\,\,\,\,\,\,\,2x=\frac{{7\pi }}{4}+2\pi k\end{array}$$   Looking at the Unit Circle, this can be simplified to: $$\displaystyle \begin{array}{c}2x=\frac{\pi }{4}+\pi k\,\,\,\,\,\,\,\,\,\,2x=\frac{{3\pi }}{4}+\pi k\\\,\,\,\,\,\,\,\,x=\frac{\pi }{8}+\frac{\pi }{2}k,\,\,\,x=\frac{{3\pi }}{8}+\frac{\pi }{2}k\end{array}$$   Looking at the Unit Circle, we can get solutions between 0 and $$2\pi$$ by adding $$\frac{{4\pi }}{8}$$, staying under $$2\pi$$: $$\displaystyle x=\left\{ \begin{array}{l}\frac{\pi }{8},\frac{{5\pi }}{8},\frac{{9\pi }}{8},\frac{{13\pi }}{8},\\\frac{{3\pi }}{8},\frac{{7\pi }}{8},\frac{{11\pi }}{8},\frac{{15\pi }}{8}\end{array} \right\}$$ $$\displaystyle \tan \left( {\frac{\theta }{2}-\frac{\pi }{3}} \right)=-1$$   $$\displaystyle \begin{array}{c}\frac{\theta }{2}-\frac{\pi }{3}=\frac{{3\pi }}{4}+\pi k\\\frac{\theta }{2}-\frac{\pi }{3}=\frac{{7\pi }}{4}+\pi k\end{array}$$   Looking at the Unit Circle, this can be simplified to: $$\displaystyle \begin{array}{c}\frac{\theta }{2}-\frac{\pi }{3}=\frac{{3\pi }}{4}+\pi k\\\frac{\theta }{2}=\left( {\frac{{3\pi }}{4}+\frac{\pi }{3}} \right)+\,\,\pi k\\\frac{\theta }{2}=\frac{{13\pi }}{{12}}+\pi k\\\theta =\frac{{13\pi }}{6}+2\pi k\end{array}$$   Looking at the Unit Circle, the only solution that will work is: $$\begin{array}{c}\frac{{13\pi }}{6}-2\pi =\frac{\pi }{6}\\\theta =\left\{ {\frac{\pi }{6}} \right\}\end{array}$$

# Factoring to Solve Trigonometric Equations

Note that sometimes we have to factor the equations to get the solutions, typically if they are trig quadratic equations. Then we set all factors to 0 to solve, making sure we test the answers to see if they work. We learned how to factor Quadratic Equations in the Solving Quadratics by Factoring and Completing the Square section.

Note that when we factor trig equations to find solutions, like we do with “regular” equations, we never just divide a factor out from each side. In doing this, we are probably “throwing away” valid solutions to the equation.

Here are some examples, both solving on the interval 0 to $$2\pi$$ and over the reals; note one of the problems is using degrees instead of radians.
You will notice in the last problem that the answer in the left column ($$\displaystyle \theta = \frac{{\pi k}}{2}$$) has to be “thrown out”, because of our domain restriction for cot (it falls on an asymptote). Thus this answer is an extraneous solution:

 Factoring to Solve Trig Equations $$\displaystyle \begin{array}{c}4\sin x\cos x=2\sin x\\\text{Interval }(0,2\pi )\end{array}$$   $$\displaystyle \begin{array}{c}4\sin x\cos x-2\sin x=0\\2\sin x\left( {2\cos x-1} \right)=0\\2\sin x=0\,\,\,\,\,\,\,\,\,\,2\cos x-1=0\\\sin x=0\,\,\,\,\,\,\,\,\,\,\cos x=\frac{1}{2}\\x=0,\,\,\pi \,\,\,\,\,\,\,\,x=\frac{\pi }{3},\,\frac{{5\pi }}{3}\\\\\,x=\left\{ {0,\pi ,\frac{\pi }{3},\frac{{5\pi }}{3}} \right\}\end{array}$$ $$\displaystyle \begin{array}{c}4{{\cos }^{4}}\theta -7{{\cos }^{2}}\theta +3=0\\\text{General Solutions (in degrees)}\end{array}$$   $$\displaystyle \begin{array}{c}\left( {4{{{\cos }}^{2}}\theta -3} \right)\left( {{{{\cos }}^{2}}\theta -1} \right)=0\\\,\sqrt{{{{{\cos }}^{2}}\theta }}=\sqrt{{\frac{3}{4}}}\,\,\,\,\,\,\,\,\,\,\,\,\sqrt{{{{{\cos }}^{2}}\theta }}=\sqrt{1}\\\,\cos \theta =\pm \frac{{\sqrt{3}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \theta =\pm 1\,\\\text{(Add }180{}^\circ k\text{ instead of }360{}^\circ k\\\text{ because of }\pm )\\\,\,\,\theta =30{}^\circ +180{}^\circ k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta =180{}^\circ k\\\,\theta =330{}^\circ +180{}^\circ k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\left\{ \begin{array}{l}\theta |\theta =30{}^\circ +180{}^\circ k,\,\,\theta =330{}^\circ +180{}^\circ k,\\\,\,\,\,\theta =180{}^\circ k\end{array} \right\}\end{array}$$ $$\displaystyle \begin{array}{c}2{{\sec }^{2}}x-3\sec x=2\\\text{General Solutions}\end{array}$$   $$\displaystyle \begin{array}{c}2{{\sec }^{2}}x-3\sec x-2=0\\\left( {2\sec x+1} \right)\left( {\sec x-2} \right)=0\\\,\,\,\sec x=-\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec x=2\\\text{(no solution)}\,\,\,\,\,\,\,\,\,\,\,\,\left( {\cos x=\frac{1}{2}} \right)\\\\\left\{ {x|x=\frac{\pi }{3}+2\pi k,x=\frac{{5\pi }}{3}+2\pi k\,} \right\}\end{array}$$ $$\displaystyle \begin{array}{c}{{\tan }^{4}}x-4{{\tan }^{2}}x+3=0\\\text{General Solutions}\end{array}$$   $$\displaystyle \begin{array}{c}\left( {{{{\tan }}^{2}}x-3} \right)\left( {{{{\tan }}^{2}}x-1} \right)=0\\{{\tan }^{2}}x-3=0\,\,\,\,\,\,\,{{\tan }^{2}}x-1=0\\\sqrt{{{{{\tan }}^{2}}x}}=\sqrt{3}\,\,\,\,\,\,\,\sqrt{{{{{\tan }}^{2}}x}}=\sqrt{1}\\\tan x=\pm \sqrt{3}\,\,\,\,\,\,\,\,\tan x=\pm \sqrt{1}\\x=\frac{\pi }{3}+\pi k\,\,\,\,\,\,\,\,\,\,x=\frac{\pi }{4}+\pi k\\x=\frac{{2\pi }}{3}+\pi k\,\,\,\,\,\,\,\,\,\,x=\frac{{3\pi }}{4}+\pi k\\\\\left\{ \begin{array}{l}x|x=\frac{\pi }{3}+\pi k,\,\,\,x=\frac{{2\pi }}{3}+\pi k,\\\,\,\,\,x=\frac{\pi }{4}+\pi k,\,\,\,x=\frac{{3\pi }}{4}+\pi k\end{array} \right\}\end{array}$$ $$\displaystyle \begin{array}{c}3{{\cot }^{2}}x+3\cot x-\sqrt{3}\cot x=\sqrt{3}\\\text{Interval }(0,2\pi )\end{array}$$   $$\displaystyle \begin{array}{c}3{{\cot }^{2}}x+3\cot x-\sqrt{3}\cot x-\sqrt{3}=0\\3\cot x\left( {\cot x+1} \right)-\sqrt{3}\left( {\cot x+1} \right)=0\\\left( {\cot x+1} \right)\left( {3\cot x-\sqrt{3}} \right)=0\\\,\,\cot x=-1\,\,\,\,\,\,\,\,\,\,\,\cot x=\frac{{\sqrt{3}}}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan x=-1\,\,\,\,\,\,\,\,\,\tan x=\frac{3}{{\sqrt{3}}}=\sqrt{3}\\\,\,\,x=\frac{{3\pi }}{4},\,\,\frac{{7\pi }}{4}\,\,\,\,\,\,\,\,\,\,\,x=\frac{\pi }{3},\,\,\frac{{4\pi }}{3}\\\\x=\left\{ {\frac{{3\pi }}{4},\frac{{7\pi }}{4},\frac{\pi }{3},\frac{{4\pi }}{3}} \right\}\end{array}$$ $$\displaystyle \begin{array}{c}\sqrt{3}\sin \left( {2\theta } \right)\cot \left( {2\theta } \right)-\sin \left( {2\theta } \right)=0\\\text{Interval }(0,2\pi )\end{array}$$   $$\displaystyle \begin{array}{c}\sin \left( {2\theta } \right)\left( {\sqrt{3}\cot \left( {2\theta } \right)-1} \right)=0\\\sin \left( {2\theta } \right)=0\,\,\,\,\,\,\,\,\,\,\,\,\cot \left( {2\theta } \right)=\frac{1}{{\sqrt{3}}}\\2\theta =\pi k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\theta =\frac{\pi }{3}+\pi k\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\theta =\frac{{4\pi }}{3}+\pi k\\\,\,\,\theta =\frac{{\pi k}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta =\frac{\pi }{6}+\frac{\pi }{2}k\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta =\frac{{4\pi }}{6}+\frac{\pi }{2}k\end{array}$$ Because of the domain restriction for cot (where its asymptotes are), and noting that $$\cot \left[ {2\left( {\frac{{\pi k}}{2}} \right)} \right]=\cot \left( {\pi k} \right)$$ is undefined, we have to eliminate $$\frac{{\pi k}}{2}$$. Solutions are: $$\theta =\left\{ {\frac{\pi }{6},\frac{{2\pi }}{3},\frac{{7\pi }}{6},\frac{{5\pi }}{3}} \right\}$$

# Solving Trigonometric Equations Using a Calculator

We already used a calculator to find inverse trig functions in a calculator here in the Inverse Trigonometric Functions section. When solving trig equations, however, it’s a little more complicated, since typically we’ll have multiple solutions.

We can do this using just a scientific calculator, or graphing the functions and finding intersections with a graphing calculator (usually easier). Don’t forget to change to the appropriate mode (radians or degrees) using DRG on a TI scientific calculator, or mode on a TI graphing calculator.

If just using a scientific calculator, here are some rules for solving trig problems in the intervals $$\left[ {0,2\pi } \right)$$ or $$\left( {-\infty ,\infty } \right)$$ in radians (substitute 180° for  if using degrees). Remember these rules, which make sense if you look at the trig functions on the Unit Circle. Remember that $$k$$ is any integer, negative, 0, or positive.

• For $$\displaystyle \sin \theta =A,\,\theta ={{\sin }^{{-1}}}A\,\,\,\left( {+\,2\pi k} \right)$$ and also $$\displaystyle \theta =\pi -{{\sin }^{{-1}}}A\,\,\left( {+\,2\pi k} \right)$$. (For $$\csc \theta =A$$, use $$\displaystyle {{\sin }^{{-1}}}\left( {\frac{1}{A}} \right)$$). For example, in the interval $$\left[ {0,2\pi } \right)$$, for $$\displaystyle \sin \theta =-.5,\,\,\theta ={{\sin }^{{-1}}}\left( {-.5\,} \right)\approx -.524\,\,\left( {-\frac{\pi }{6}\,\,\,\text{or}\,\,\frac{{11\pi }}{6}\,} \right)$$, and also $$\displaystyle \theta =\pi -{{\sin }^{{-1}}}\left( {-.5} \right)\,=\pi -\frac{{11\pi }}{6}=\,\,\frac{{5\pi }}{6}$$.
• For $$\displaystyle \cos \theta =A,\,\theta ={{\cos }^{{-1}}}A\,\,\,\left( {+\,2\pi k} \right)$$, and also $$\displaystyle \theta =2\pi -{{\cos }^{{-1}}}A\,\,\left( {+\,2\pi k} \right)$$, which is the same as $$\displaystyle -{{\cos }^{{-1}}}A\,\,\left( {+\,2\pi k} \right)$$. (For $$\sec \theta =A$$, use $$\displaystyle {{\cos }^{{-1}}}\left( {\frac{1}{A}} \right)$$ ). For example, in the interval $$\left[ {0,2\pi } \right)$$, for $$\displaystyle \cos \theta =.5,\,\,\theta ={{\cos }^{{-1}}}\left( {.5} \right)\,\approx 1.047\,\,\left( {\frac{\pi }{3}} \right)$$, and also $$\displaystyle \theta =2\pi -{{\cos }^{{-1}}}\left( {.5} \right)\,\approx 5.236\,\,\left( {\frac{{5\pi }}{3}} \right)$$.
• For $$\displaystyle \tan \theta =A,\,\,\theta ={{\tan }^{{-1}}}A\,\,\,\left( {+\,\pi k} \right)$$; this will find all the solutions. (For $$\cot \theta =A$$, use $$\displaystyle {{\tan }^{{-1}}}\left( {\frac{1}{A}} \right)$$). For example, in the interval $$\left[ {0,2\pi } \right)$$, for $$\displaystyle \tan \theta =1,\,\,\theta ={{\tan }^{{-1}}}\left( 1 \right)\,\approx .785\,\,\left( {\frac{\pi }{4}} \right)$$, and also $$\displaystyle \theta ={{\tan }^{{-1}}}\left( 1 \right)+\pi \,\approx 3.927\,\,\left( {\frac{{5\pi }}{4}} \right)$$.

Remember that when the coefficient of the argument of the inverse trig functions isn’t 1, we need to divide the $$+\,2\pi k$$ or $$+\,\pi k$$ by this coefficient, since the period changes (see examples below).

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If you have access to a graphing calculator, it’s usually easier to solve trig equations. We can put the left-hand part of the equation in $${{Y}_{1}}$$ and the right-hand part of the equation in $${{Y}_{2}}$$ and solve for the intersection(s) between 0 and $$2\pi$$. For solving over the real (general solutions), you can put the period of the trig function and then add the appropriate factors of $$\pi k$$, $$2\pi k$$, or whatever the period of the function is.

Remember for the reciprocal functions, take the reciprocal of what’s on the right hand side, and use the regular trig functions.

Typically, the default mode is radian mode, unless problem says “degrees”. When finding the intersection on the graphing calculator, use the TRACE and arrow buttons to move the cursor closer to the points of intersection, since the Intersect function will use the closest intersection.

For intervals of $$\left[ {0,2\pi } \right)$$, use Xmin = 0, and Xmax = $$2\pi$$. For general solutions (over the reals), use Xmin = 0 and Xmax = the period (such as $$\displaystyle \frac{2\pi }{5}$$ when you have $$\sin \left( {5x} \right)$$, for example) for general solutions.

Also remember that $${{\left( \cos \theta \right)}^{2}}$$ is written as $${{\cos }^{2}}\theta$$, and we can put it in the graphing calculator as $$\boldsymbol{\cos {{\left( x \right)}^{2}}}$$ or $$\boldsymbol {{{\left( {\cos \left( x \right)} \right)}^{2}}}$$.

Here are some examples using both types of calendars:

 Solving Trig Problems with a Calculator $$\displaystyle \sin x=\frac{3}{4}\text{ over Reals}$$ (in degrees) Without Graphing: To get the second value of arcsin, subtract the first from 180 or $$\pi$$. With Graphing: Use 0 for Xmin and 360 for Xmax: Add $$360{}^\circ k$$ to answers: $$\displaystyle \left\{ \begin{array}{l}x|\,x=48.6+360k\\\,\,\,\,\,\,\,\,\,\,=131.4+360k\end{array} \right\}$$ $$\displaystyle \tan \left( {2\theta } \right)=-\frac{1}{{\sqrt{6}}}\text{ on }\left[ {0,2\pi } \right)$$ (in radians)   Without Graphing: To get the second value of arctan, add 180 or $$\pi$$. For this problem, add $$\displaystyle \frac{\pi }{2}$$ (the period) to get all the values between 0 and $$2\pi$$. With Graphing: Use 0 for Xmin and $$2\pi$$ for Xmax:   Get all intersections: $$\left\{ \begin{array}{l}\theta |\theta =1.377,\,2.948\,\\\,\,\,\,\,\,\,\,=4.519,\,\,6.089\end{array} \right\}$$ $$\displaystyle \cos \left( {3\theta -1} \right)-.8=0\text{ over Reals}$$ Without Graphing: Work backwards to get the value of $$x$$. To get the second value of arccos, we can take the negative cosine (same as subtracting the first from 360 or $$2\pi$$). For the answers .548 and .119, add $$\displaystyle \frac{{2\pi }}{3}k$$ (period) to get all the solutions. With Graphing: (Easier!) Note that we use 0 for Xmin and $$\displaystyle \frac{{2\pi }}{3}$$ for Xmax since $$\displaystyle \frac{{2\pi }}{3}$$ is the period: Get both intersections and add factors of $$\displaystyle \frac{{2\pi }}{3}$$ (the period): \displaystyle \left\{ \begin{align}\theta |\,\theta &=.119+\frac{{2\pi k}}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,&=\,.548+\frac{{2\pi k}}{3}\end{align} \right\} $$\displaystyle \cot \left( {\frac{2}{5}x-\frac{\pi }{4}} \right)\text{ =10 over Reals}$$   Without Graphing: Work backwards to get the value of $$x$$, using arctan of $$\displaystyle \frac{1}{{10}}$$. For the answer 2.213, add $$\displaystyle \frac{{5\pi }}{2}k$$ (period) to get all the solutions. With Graphing: Note that we use 0 for Xmin and $$\displaystyle \frac{{\,\pi }}{{\left( {\frac{2}{5}} \right)}}=\frac{{5\pi }}{2}$$ for Xmax since $$\displaystyle \frac{{5\pi }}{2}$$ is the period:   Add factors of $$\displaystyle \frac{{5\pi }}{2}$$ (the period): $$\displaystyle \left\{ {x|\,x=2.213+\frac{{5\pi k}}{2}} \right\}$$

# Solving Trig Systems of Equations

We learned how to solve systems of more complicated equations here in the Systems of non-Linear Equations section.

Again, we can use either Substitution or Elimination, depending on what’s easier. Once we get the initial solution(s), we’ll need to plug in to get the other variable.

Here are some examples of Solving Systems with Trig Equations; solve over the reals:

 Solving Trig Systems of Equations $$\displaystyle \begin{array}{c}y-\cos \left( x \right)=0\\y=\sin \left( x \right)\end{array}$$   $$\displaystyle \begin{array}{c}\sin x-\cos x=0\\\sin x=\cos x\\\frac{{\sin x}}{{\cos x}}=1\\\,\tan x=1\end{array}$$              $$\displaystyle \begin{array}{c}x=\frac{\pi }{4}+2\pi k\,\,\,\,\,\,\,\,x=\frac{{3\pi }}{4}+2\pi k\\y=\sin \left( {\frac{\pi }{4}} \right)=\frac{{\sqrt{2}}}{2}\,\,\,\,\,\,\,\,\,y=\sin \left( {\frac{{3\pi }}{4}} \right)=-\frac{{\sqrt{2}}}{2}\\\,\left( {\frac{\pi }{4}+2\pi k,\,\,\frac{{\sqrt{2}}}{2}} \right)\,\,\,\,\,\,\,\,\,\left( {\frac{{3\pi }}{4}+2\pi k,\,\,-\frac{{\sqrt{2}}}{2}} \right)\end{array}$$ We’ll use substitution to get one equation, only with $$x$$’s. Since both sin and cos go on and on in the $$x$$ direction, we have to include “$$\displaystyle +2\pi k$$” to our solutions.   (We have to be careful to separate the solutions we get to $$\displaystyle x=\frac{\pi }{4}+2\pi k$$ and $$\displaystyle x=\frac{{3\pi }}{4}+2\pi k$$, even though with tan, we can typically use $$\displaystyle x=\frac{\pi }{4}+\pi k$$ for this solution. The reason we have to separate our answers like this is because we’ll have different answers for $$y$$ for the two different places on the unit circle.)   Once we get the solutions for $$x$$, we plug these back in either equation to get our $$y$$ values. We see that the solutions are $$\displaystyle \left( {\frac{\pi }{4}+2\pi k,\frac{{\sqrt{2}}}{2}} \right)$$ and $$\displaystyle \left( {\frac{{3\pi }}{4}+2\pi k,-\frac{{\sqrt{2}}}{2}} \right)$$. $$\displaystyle \begin{array}{c}\cos \left( x \right)+y=3\\y={{x}^{2}}\end{array}$$   Put in graphing calculator:   $$\displaystyle \begin{array}{l}{{Y}_{1}}=3-\cos \left( x \right)\\{{Y}_{2}}={{x}^{2}}\end{array}$$ Let’s do this one on our graphing calculator (make sure your calculator is in radians). Solve for $$y$$ in both cases, graph, and find the intersection. I made my window between $$-2\pi$$ and $$2\pi$$ for $$x$$.   Use the intersect feature on the calculator (2nd trace, 5, enter, enter, enter) to find the intersection. (Use trace and arrow keys to get close to each intersection before using intersect). If we zoom out, we can see there are only two solutions. We see that the only solutions are $$\left( {-1.795,3.222} \right)$$ and $$\left( {1.795,3.222} \right)$$.

# Solving Trig Inequalities

Sometimes you might be asked to solve a Trig Inequality. There are links to many other types of Inequalities here (we saw one of these above).

We can either solve these inequalities graphically or algebraically; let’s try one of each. Note that you can also solve these on your graphing calculator, using the Intersect feature, and then see where the inequalities “work”:

 Solving Trig Inequalities Graphically Solving Trig Inequalities Algebraically $$-6\left| {\sin \theta } \right|\le -3\,\,\,\,\,\,\text{over reals}$$ $$\displaystyle \left| {\sin \theta } \right|\ge \frac{1}{2}\,\,\,\,\,\,\text{(switch signs!)}$$ Graph both sides of the inequality to see where $$\left| {\sin \left( x \right)} \right|$$ is greater than or equal to $$\displaystyle \frac{1}{2}$$. Just graph between 0 and $$2\pi$$ to see the points of intersection. We want above (including) $$\displaystyle \frac{1}{2}$$, because of the $$\ge$$: We see the solution is: $$\displaystyle \left( {\frac{\pi }{6}+2\pi k,\frac{{5\pi }}{6}+2\pi k} \right)\cup \left( {\frac{{7\pi }}{6}+2\pi k,\frac{{11\pi }}{6}+2\pi k} \right)$$ Set everything to 0 and factor, so we can get our critical values:   $$\displaystyle 3{{\cot }^{2}}x+3\cot x-\sqrt{3}\cot x<\sqrt{3};\,\,\,\,\,\,0\le x\le 2\pi$$ $$\displaystyle \begin{array}{c}3{{\cot }^{2}}x+3\cot x-\sqrt{3}\cot x-\sqrt{3}<0\\3\cot x\left( {\cot x+1} \right)-\sqrt{3}\left( {\cot x+1} \right)<0\\\left( {\cot x+1} \right)\left( {3\cot x-\sqrt{3}} \right)<0\end{array}$$ $$\displaystyle \,\cot x=-1\,\,\,\,(x=\frac{{3\pi }}{4},\,\,\frac{{7\pi }}{4})\,\,\,\,\,\,\,\,\,\,\,\,\cot x=\frac{{\sqrt{3}}}{3}\,\,\,\,(x=\frac{\pi }{3},\,\,\frac{{4\pi }}{3})$$   Draw sign chart with critical values $$\displaystyle \frac{{3\pi }}{4},\frac{{7\pi }}{4},\frac{\pi }{3},\,\frac{{4\pi }}{3},$$ and endpoints 0 and $$2\pi$$, since we’re just going between 0 and $$2\pi$$. Use open circles for the critical values since we have a $$<$$. Then check each interval with a sample value in the last inequality above and see if we get a positive or negative value. We need to take values where the function is defined (avoid asymptotes), so let’s try $$\displaystyle \frac{\pi }{4}$$ for the interval less than $$\displaystyle \frac{\pi }{3}$$ for example: $$\displaystyle \left( {\cot \frac{\pi }{4}+1} \right)\left( {3\cot \frac{\pi }{4}-\sqrt{3}} \right)\approx 2.5$$, which is positive: We want the negative intervals, not including the critical values. We see the solution is: $$\displaystyle \left( {\frac{\pi }{3},\frac{{3\pi }}{4}} \right)\cup \left( {\frac{{4\pi }}{3},\frac{{7\pi }}{4}} \right)$$.

Practice these problems, and practice, practice, practice!

Hit Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

For Practice: Use the Mathway widget below to try a Trig Solving problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve for x to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Trigonometric Identities – you’re ready!

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text: Non-perturbative analysis of the spectrum of meson resonances in an ultraviolet-complete composite-Higgs model Auteur(s): Bizot N., Frigerio M., Knecht Marc, Kneur J.-L. (Article) Publié: Physical Review D, vol. 95 p.075006 (2017) Texte intégral en Openaccess : Ref HAL: hal-01775256_v1 Ref Arxiv: 1610.09293 DOI: 10.1103/PhysRevD.95.075006 Ref. & Cit.: NASA ADS Exporter : BibTex | endNote 7 citations Résumé: We consider a vector-like gauge theory of fermions that confines at the multi-TeV scale, and that realizes the Higgs particle as a composite Goldstone boson. The weak interactions are embedded in the unbroken subgroup Sp(4) of a spontaneously broken SU (4) flavour group. The meson resonances appear as poles in the two-point correlators of fermion bilinears, and include the Goldstone bosons plus a massive pseudoscalar η , as well as scalars, vectors and axial vectors. We compute the mass spectrum of these mesons, as well as their decay constants, in the chiral limit, in the approximation where the hypercolour Sp(2N) dynamics is described by four-fermion operators, ` a la Nambu-Jona Lasinio. By resumming the leading diagrams in the 1/N expansion, we find that the spin-one states lie beyond the LHC reach, while spin-zero electroweak-singlet states may be as light as the Goldstone-boson decay constant, f ∼ 1 TeV. We also confront our results with a set of available spectral sum rules. In order to supply composite top-quark partners, the theory contains additional fermions carrying both hypercolour and ordinary colour, with an associated flavour symmetry-breaking pattern SU (6)/SO(6). We identify and analyse several non-trivial features of the complete two-sector gauge theory: the 't Hooft anomaly matching conditions; the higher-dimension operator which incorporates the effects of the hypercolour axial-singlet anomaly; the coupled mass-gap equations; the mixing between the singlet mesons of the two sectors, resulting in an extra Goldstone boson η0, and novel spectral sum rules. Assuming that the strength of the four-fermion interaction is the same in the two sectors, we find that the coloured vector and scalar mesons have masses 4f , while the masses of coloured pseudo-Goldstone bosons, induced by gluon loops, are 1.5f. We discuss the scaling of the meson masses with the values of N , of the four-fermion couplings, and of a possible fermion mass. Commentaires: 65 pages, 25 figures

format: plain


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# zbMATH — the first resource for mathematics

On the effectiveness in a problem of nonlinear prognosis and filtration. (Russian. English summary) Zbl 0553.60046
Let $$\zeta =\phi (\eta),\xi_ t=\phi_ t(\eta_ t)$$, $$t\in T$$, where $$\phi(\cdot)$$, $$\phi_ t(\cdot)\in L_ 2(d\Phi)$$ with the standard normal distribution $$\Phi$$ and $$\eta$$, $$\eta_ t$$, $$t\in T$$, is a Gaussian system of random variables with parameters in (0,1). The question of finding the best (in the mean square sense) linear and nonlinear estimations of a random variable $$\zeta$$ (when the $$\eta_ t$$, $$t\in T$$, system is observed) and the character change of the $$\bar D/\tilde D$$ relation are studied, where $$\bar D$$ is the mean square error of linear estimation, and $$\tilde D$$ is the mean square error of nonlinear estimation.
##### MSC:
 60G35 Signal detection and filtering (aspects of stochastic processes) 60G15 Gaussian processes 93E10 Estimation and detection in stochastic control theory


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# Evaluate $\int\frac{dx}{\sin(x+a)\sin(x+b)}$ [duplicate]

This question already has an answer here:

Please help me evaluate: $$\int\frac{dx}{\sin(x+a)\sin(x+b)}$$

## marked as duplicate by Andrew D. Hwang, Laurent Duval, Juniven, Dragonemperor42, Michael HoppeApr 23 '17 at 17:40

The given integral is: $$\int\frac{dx}{\sin(x+a)\sin(x+b)}$$

The given integral can write:

$$\int\frac{dx}{\sin(x+a)\sin(x+b)}=\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}$$

We substition $$\frac{\sin(x+a)}{\sin(x+b)}=t$$

By the substition of the above have:

$$\frac{dx}{\sin^2(x+a)}=\frac{dt}{\sin(a-b)}$$

Now have:

$$\int\frac{dx}{\sin(x+a)\sin(x+b)}=\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}=\frac{1}{\sin(a-b)}\int\frac{dt}{t}=\frac{1}{\sin(a-b)}\ln |t|=\frac{1}{\sin(a-b)}\ln|\frac{\sin(x+a)}{\sin(x+b)}|+C$$

• This is supposed to be a homework question, so I presume you should leave the proposer with some room to do it himself... – Rijul Saini Sep 4 '12 at 20:14

Hint: Multiply and divide by $\sin(b-a)$.

Further Hint:

Write $\sin(b-a) = \sin((x-a)-(x-b))$


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# [metapost] Question: [MPLib, MetaFun, ConTeXt, LuaTeX] metapost logging and message primitive?

Hans Hagen pragma at wxs.nl
Tue Feb 26 11:44:09 CET 2013

On 2/26/2013 11:14 AM, Colin Fraizer wrote:
> Perhaps this is a simpler question:
>
> Is there a way for me to see exactly the code ConTeXt passes to
> MPLIB so I can put that in a ".mp" file and run it with METAPOST?
>
> That would let me look at the logs and see where things go wrong.

see attached ... runs ok on my machine with latest context beta and
latest luatex beta ... the tracker tells context to create an mp file too

> %% ----------------------------------------------------------------------
> \loggingall

that will give pretty large files as there's a lot going on at the tex end

Hans

ps. given mp development and the fact that context adapts to the latest
mp and luatex it's best to run the minimals from the contextgarden;
these can be installed alongside texlive

-----------------------------------------------------------------
Hans Hagen | PRAGMA ADE
Ridderstraat 27 | 8061 GH Hasselt | The Netherlands
tel: 038 477 53 69 | voip: 087 875 68 74 | www.pragma-ade.com
| www.pragma-pod.nl
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\enabletrackers[metapost.graphics]

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fill fullcircle scaled 4cm withcolor blue; label.top(btex \color[white]{\bf This is a test.} etex, origin);
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# Also in the Article

Corpus and document similarity
This protocol is extracted from research article:
The public and legislative impact of hyperconcentrated topic news

Procedure

We estimate the similarity of a corpus of documents as the median of its pairwise document similarities, using all $(n2)$ combinations from the corpus. To estimate similarity between two documents, we adopted doc2vec (12), a well-known tool that generates a vector representation (called a “paragraph vector”) of a document. Specifically, we used a standard doc2vec model (30), trained on each domain corpus, to compute a vector for each document in our corpus. We defined the pairwise similarity of two documents as the cosine similarity of their respective document vectors (31).

Whereas we do not in general deny that high median similarities can occur in annual corpora with low news volume (see fig. S1), we found that legislative activity tends to correlate with periods in which news volume and median similarity are simultaneously high. We therefore employ a threshold whereby the similarity of an annual corpus is considered to be zero if it contains less than c% of the articles from the respective domain corpus. We use a threshold of c = 5% in this paper.

We note that since cosine similarity has a range of [−1, 1], and our models are learnt on datasets that discuss a common topic, the variation in similarities we obtain is relatively small compared to a metric with a larger range.

Despite this conservative choice, we demonstrate consistent G causality with legislation (table S2). We note that stronger significance may be obtained if we were to use similarity measures with larger ranges. However, the results obtained with our conservative approach inspire confidence in the validity of our hypothesis.

Note: The content above has been extracted from a research article, so it may not display correctly.

Q&A


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# Convergence of a series

#### chwala

Gold Member
1. The problem statement, all variables and given/known data
Determine whether the series $\frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$ converges or not

2. Relevant equations

3. The attempt at a solution
looking at $1/sin (n)$ by comparison,
$1/n^2=1+1/4+1/9+1/16+...$ converges for $n≥1$
for $n≥1$
implying that ${sin (n)}≤n$
$1/2sin (n) ≤1/{n}$ converges, up to this point i have been trying to look at the trig. term in the series....
again by comparison,...

Related Calculus and Beyond Homework News on Phys.org

#### tnich

Homework Helper
1. The problem statement, all variables and given/known data
Determine whether the series $\frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$ converges or not

2. Relevant equations

3. The attempt at a solution
looking at $1/sin (n)$ by comparison,
$1/n^2=1+1/4+1/9+1/16+...$ converges for $n≥1$
for $n≥1$
implying that ${sin (n)}≤n$
$1/2sin (n) ≤1/{n}$ converges, up to this point i have been trying to look at the trig. term in the series....
again by comparison,...
I think the first thing you need to do is eliminate the trig term by finding a bounding series. What series $f(n)$ can you come up that is just a little bit larger than $\frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$ and has no trig term?

#### Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
1. The problem statement, all variables and given/known data
Determine whether the series $\frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$ converges or not

2. Relevant equations

3. The attempt at a solution
looking at $1/sin (n)$ by comparison,
$1/n^2=1+1/4+1/9+1/16+...$ converges for $n≥1$
for $n≥1$
implying that ${sin (n)}≤n$
$1/2sin (n) ≤1/{n}$ converges, up to this point i have been trying to look at the trig. term in the series....
again by comparison,...
Look at
$$\text{numerator} = (n^3 +3n)^{1/2} = n^{3/2} \left( 1 + \frac 3 n \right)^{1/2}$$
and
$$\text{denominator} = 5n^3 + 3 n^2 + 2 \sin(n) = 5 n^3 \left( 1 + \frac{3}{5n} + \frac{2 \sin(n)}{5n^3} \right).$$
The numerator is $< 2 \, n^{3/2}$, and for any small $\epsilon > 0$ we can find $N> 0$ so that $n > N$ implies the denominator is $> 5n^3 (1 - \epsilon).$

Last edited:

#### chwala

Gold Member
Look at
$$\text{numerator} = (n^3 +3n)^{1/2} = n^{3/2} \left( 1 + \frac 3 n \right)^{1/2}$$
and
$$\text{denominator} = 5n^3 + 3 n^2 + 2 \sin(n) = 5 n^3 \left( 1 + \frac{3}{5n} + \frac{2 \sin(n)}{5n^3} \right).$$
The numerator is $< 2 \, n^{3/2}$, and for any small $\epsilon > 0$ we can find $N> 0$ so that $n > N$ implies the denominator is $> 5n^3 (1 - \epsilon).$
let me post my attempt..

#### chwala

Gold Member
there look at my working attached........

#### Attachments

• 14.9 KB Views: 77

#### chwala

Gold Member
since the series $1/n^3$ converges for the p-series p greater than 1, it suffices to say that our series converges though it fails for the limit comparison test....my thoughts

#### Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
there look at my working attached........
This attachment shows up on my laptop computer, but opens empty and blank on my i-phone. You should avoid using tools that require the reader to use particular media to read your posts. Avoid WORD; use LaTeX instead. It is faster, easier and looks better. Now is the time for you to start learning it; you can install a complete system on your computer for $0 (for typing up your assignments), and it is built-in in this Forum as well. #### chwala Gold Member This attachment shows up on my laptop computer, but opens empty and blank on my i-phone. You should avoid using tools that require the reader to use particular media to read your posts. Avoid WORD; use LaTeX instead. It is faster, easier and looks better. Now is the time for you to start learning it; you can install a complete system on your computer for$0 (for typing up your assignments), and it is built-in in this Forum as well.
ok sir, allow me to repost in latex....doing so in a few hours let me finish with a class...

#### chwala

Gold Member
our problem $\sum_{n=1}^\infty$ $\frac {(n^2+3n)^{1/2}} {5n^3+3n^2+ 2 sin n}\$
*give me a few more hours finishing with a class*
$sum_{n=1}^\infty$ ${\frac {3}{n^2}+1}$
GIVE ME TIMEAM IN CLASS

Last edited:

#### Delta2

Homework Helper
Gold Member
I also cant read your work in word document (yeap living in 2018 and haven't installed Microsoft Word :D)…

But , following the hints from @Ray Vickson, the final sequence $b_n$ which we ll use for the direct comparison test is $b_n=\frac{2n^{3/2}}{5\epsilon n^3}=\frac{2}{5\epsilon}\frac{1}{n^{3/2}}$ so your final conclusion should be "because the series of $b_n$ converges (p-series with p=3/2>1) and $|a_n|<|b_n|$ for n>N for some positive $\epsilon$ and $N$, from direct comparison test it follows that the series of $a_n$ converges."

#### chwala

Gold Member
I also cant read your work in word document (yeap living in 2018 and haven't installed Microsoft Word :D)…

But , following the hints from @Ray Vickson, the final sequence $b_n$ which we ll use for the direct comparison test is $b_n=\frac{2n^{3/2}}{5\epsilon n^3}=\frac{2}{5\epsilon}\frac{1}{n^{3/2}}$ so your final conclusion should be "because the series of $b_n$ converges (p-series with p=3/2>1) and $|a_n|<|b_n|$ for n>N for some positive $\epsilon$ and $N$, from direct comparison test it follows that the series of $a_n$ converges."
kindly let's be patient as i am reposting the solution using latex, i am learning a few latex terms.....give me time...

#### Mark44

Mentor
@chwala, the work you showed in the Word document is wrong right from the start.
You have
$$\sum_{n = 1}^\infty \frac{\sqrt{n^2 + 3n}}{5n^3 + 3n^2 + 2\sin n} = \sum_{n = 1}^\infty \frac {\frac{\sqrt{n(n^2 + 3)}}{n^3}}{5 + \frac 3 n + \frac{2\sin n}{n^3}}$$

In post #1 you have $(n^3 + 3n)^{1/2}$ in the numerator, but above you have $\sqrt{n^2 + 3n}$ on the left side. That's a relatively minor typo, but confusing for someone who might not have read things carefully throughout the thread.

BTW, since you're struggling with LaTeX, here's what I typed for the stuff above, in unrendered form:
\sum_{n = 1}^\infty \frac{\sqrt{n^2 + 3n}}{5n^3 + 3n^2 + 2\sin n} = \sum_{n = 1}^\infty \frac {\frac{\sqrt{n(n^2 + 3)}}{n^3}}{5 + \frac 3 n + \frac{2\sin n}{n^3}}

One thing to keep in mind for exponents, subscripts, fraction parts, limits of integration, and possibly a few more things: if the exponent, fraction part, etc. is just a single character, you don't need to include braces around it.
For example, this is fine: x^2
But here you need braces: x^{-2}

And this is fine \frac 1 2
But here you need braces: \frac {x - 2} 3 and \frac {x - 2}{3x}

Last edited:

#### Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
kindly let's be patient as i am reposting the solution using latex, i am learning a few latex terms.....give me time...
Nobody is rushing you. If you need a few days, take a few days.

#### chwala

Gold Member
Nobody is rushing you. If you need a few days, take a few days.
thanks i have been busy now i can embark on physicsforums...

#### chwala

Gold Member
@chwala, the work you showed in the Word document is wrong right from the start.
You have
$$\sum_{n = 1}^\infty \frac{\sqrt{n^2 + 3n}}{5n^3 + 3n^2 + 2\sin n} = \sum_{n = 1}^\infty \frac {\frac{\sqrt{n(n^2 + 3)}}{n^3}}{5 + \frac 3 n + \frac{2\sin n}{n^3}}$$

In post #1 you have $(n^3 + 3n)^{1/2}$ in the numerator, but above you have $\sqrt{n^2 + 3n}$ on the left side. That's a relatively minor typo, but confusing for someone who might not have read things carefully throughout the thread.

BTW, since you're struggling with LaTeX, here's what I typed for the stuff above, in unrendered form:
\sum_{n = 1}^\infty \frac{\sqrt{n^2 + 3n}}{5n^3 + 3n^2 + 2\sin n} = \sum_{n = 1}^\infty \frac {\frac{\sqrt{n(n^2 + 3)}}{n^3}}{5 + \frac 3 n + \frac{2\sin n}{n^3}}

One thing to keep in mind for exponents, subscripts, fraction parts, limits of integration, and possibly a few more things: if the exponent, fraction part, etc. is just a single character, you don't need to include braces around it.
For example, this is fine: x^2
But here you need braces: x^{-2}

And this is fine \frac 1 2
But here you need braces: \frac {x - 2} 3 and \frac {x - 2}{3x}
noted, let me retype my whole solution using latex...

#### Mark44

Mentor
While you're figuring out LaTeX, something you should consider is that
$$\sum_{n = 1}^\infty \frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 + 2} \le \sum_{n = 1}^\infty \frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 + 2\sin(n)} \le \sum_{n = 1}^\infty\frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 - 2}$$
If you can show that the outer summations are convergent, you'll be able to say something about the series you're working with.

#### chwala

Gold Member
am back guys...was on vacation, let me embark on this again...

#### chwala

Gold Member
While you're figuring out LaTeX, something you should consider is that
$$\sum_{n = 1}^\infty \frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 + 2} \le \sum_{n = 1}^\infty \frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 + 2\sin(n)} \le \sum_{n = 1}^\infty\frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 - 2}$$
If you can show that the outer summations are convergent, you'll be able to say something about the series you're working with.
bet its easy to show by comparison method.

#### chwala

Gold Member
Allow me to look at this post again, i hope its not closed. I have been busy lately...

#### SammyS

Staff Emeritus
Science Advisor
Homework Helper
Gold Member
Allow me to look at this post again, i hope its not closed. I have been busy lately...
It looks to me like it's open.

#### Mark44

Mentor
Allow me to look at this post again, i hope its not closed. I have been busy lately...
Yes, it's open, although 5+ months seems a long time to figure out this problem.

#### chwala

Gold Member
Let me look at this again, i need to refresh on convergence, my apologies

#### chwala

Gold Member
Yes, it's open, although 5+ months seems a long time to figure out this problem.
can you give me more insight Mark? i am stuck in this rumble...

#### Mark44

Mentor
From post #9, of July 18:
chwala said:
our problem $\sum_{n=1}^\infty$ $\frac {(n^2+3n)^{1/2}} {5n^3+3n^2+ 2 sin n}\$
The dominant term in the numerator is n, and the dominant term in the denominator is $5n^3$. Do you know the Limit Comparison Test?

#### chwala

Gold Member
ok let me look at this using comparison test

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# 11.10: Box-and-Whisker Plots

Difficulty Level: At Grade Created by: CK-12
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Mr. Hernandez gave his students a science quiz. He has organized the quiz scores in the table below.

 Student Test Score AS 89 AS 91 BC 87 CS 77 FR 72 JW 59 ML 76 MY 68 SR 83 ST 91 VS 81 ZS 73

Mr. Hernandez wants to create a graph to display the distribution of the scores. How can he do this?

In this concept, you will learn how to create and read box-and-whisker plots.

### Creating and Reading Box-and-Whisker Plots

Data is a set of numerical or non-numerical information. Data can be analyzed in many different ways. In this concept you will analyze numerical data using box-and-whisker plots.

Median is a numerical value that represents the middle term in a data set. When ordered sequentially, the value that is in the middle of the data set is the median

The range of a set of data is the difference between the largest and smallest values. The range identifies how far apart the values are in the data set.

The minimum is the smallest value in the data set.

The maximum is the largest value in the data set.

Quartiles divide data sets into four equal groups. The first quartile is the median of the lower half of the data set. The second quartile is the median of the data set. The third quartile is the median of the upper half of the data set.

One way to display data is in a box-and-whisker plot. A box plot illustrates the distribution of the data set in quartiles. A box is made around the median, with the sides of the box being the first and third quartiles. The whiskers are the lines extending outwards from either side of the box indicating the range of the data set. Box and whisker plots can be oriented vertically or horizontally.

Let's look at an example.

Create a box-and-whisker plot to display the data below.

45, 58, 34, 42, 52, 49, 50, 45, 51

First, arrange the data in order from smallest to largest.

34, 42, 45, 45, 49, 50, 51, 52, 58

Next, find the median, which is the middle number. There are nine terms, so the fifth term is the median, 49. The median will not be used in calculating the first or third quartiles

Next, find the first quartile, which is the middle number of the lower half of the data set. Since 49 is not included, there are four terms in the lower half of the set. The median of this set is between the second and third terms, 42 and 45. To find the median we add the two terms together and divide by 2.

42+45=87\begin{align*}42+45=87\end{align*}

87÷2=43.5\begin{align*}87\div 2=43.5\end{align*}

So, the first quartile is 43.5. This will be the left side of the box.

Next, find the third quartile, which is the middle number of the upper half of the data set. Since 49 is not included, there are four terms in the upper half of the set. The median of this set is between the second and third terms, 51 and 52. To find the median we add the two terms together and divide by 2.

51+52=103\begin{align*}51+52=103\end{align*}

103÷2=51.5\begin{align*}103÷2=51.5\end{align*}

So, the third quartile is 51.5. This will be the right side of the box.

Next, find the minimum and maximum of the data set. These will be the ends of the whiskers. The minimum is the smallest term in the set. In this case, it's 34. The maximum is the largest term in the set. In this case, it's 58.

Next, draw the box-and-whisker plot. Begin by drawing a number line that begins below the minimum, 34, and extends beyond the maximum, 58. In this case the number line should be drawn from 30 to 60. Then draw the box and whisker plot above the number line.

The box-and-whisker plot should look like the one below.

### Examples

#### Example 1

Earlier, you were given a problem about Mr. Hernandez and his science quiz.

Mr. Hernandez gave his class a science quiz and wants to create a graph to show the distribution of the scores. He organized his data into the table below.

 Student Test Score AS 89 AS 91 BC 87 CS 77 FR 72 JW 59 ML 76 MY 68 SR 83 ST 91 VS 81 ZS 73

How can you represent this data in a graph?

First, decide what kind of graph to make. Since you want to show the distribution of the quiz scores, a box-and-whisker plot is best.

Next, organize the data from smallest to largest.

 Student Test Score JW 59 MY 68 FR 72 ZS 73 ML 76 CS 77 VS 81 SR 83 BC 87 AS 89 AS 91 ST 91

Next, find the median, which is the middle number. There are twelve terms, so the median is between the sixth and seventh terms, 77 and 81. To find the median, add the two values together and divide by 2.

77+81=158\begin{align*}77+81=158\end{align*}

158÷2=79\begin{align*}158\div 2=79\end{align*}

The median is 79.

Next, find the first quartile, which is the middle number of the lower half of the data set. There are six terms in the lower half of the set, 59, 58, 72, 73, 76, and 77. The median of this set is between the third and fourth terms, 72 and 73. The number that is halfway between 72 and 73 is 72.5. The first quartile is 72.5, which will be the left side of the box.

Next, find the third quartile, which is the middle number of the upper half of the data set. There are six terms in the upper half of the set, 81, 83, 87, 89, 91, and 91. The median of this set is between the third and fourth terms, 87 and 89. The number that is halfway between 87 and 89 is 88. The third quartile is 88, which will be the right side of the box.

Next, find the minimum and maximum of the data set. These will be the ends of the whiskers. The minimum is the smallest term in the set. In this case, it's 59. The maximum is the largest term in the set. In this case, it's 91.

Next, draw the box-and-whisker plot. Begin by drawing a number line that begins below the minimum, 59, and extends beyond the maximum, 91. In this case, the number line should begin at 55 and go to 95. Then draw the box-and-whisker plot above the number line.

Lastly, title the graph. The title should be short and clear. In this case, title the graph "Quiz Scores."

The box-and-whisker plot should look like the one below.

#### Example 2

The data values in the table below illustrate the number of TVs sold at a department store each month for nine months. Create a box-and-whisker plot to display the data.

April May June July August September October November December
110 98 91 102 89 95 108 118 152

First, arrange the data in order from smallest to largest.

89, 91, 95, 98, 102, 108, 110, 118, 152

Next, find the median, which is the middle number. There are nine terms, so the fifth term is the median, 102. The median will not be used in calculating the first or third quartiles

Next, find the first quartile, which is the middle number of the lower half of the data set. Since 102 is not included, there are four terms in the lower half of the set, 89, 91, 95, and 98. The median of this set is between the second and third terms, 91 and 95. To find the median we add the two terms together and divide by 2.

91+95=186\begin{align*}91+95=186\end{align*}

186÷2=93\begin{align*}186÷2=93\end{align*}

So, the first quartile is 93. This will be the left side of the box.

Next, find the third quartile, which is the middle number of the upper half of the data set. Since 102 is not included, there are four terms in the upper half of the set, 108, 110, 118, and 152. The median of this set is between the second and third terms, 110 and 118. To find the median we add the two terms together and divide by 2.

110+118=228\begin{align*}110+118=228\end{align*}

228÷2=114\begin{align*}228÷2=114\end{align*}

So, the third quartile is 114. This will be the right side of the box.

Next, find the minimum and maximum of the data set. These will be the ends of the whiskers. The minimum is the smallest term in the set. In this case, it's 89. The maximum is the largest term in the set. In this case, it's 152.

Next, draw the box-and-whisker plot. Begin by drawing a number line that begins below the minimum, 89, and extends beyond the maximum, 152. In this case, the number line should begin at 80 and go to 160. Then draw the box-and-whisker plot above the number line.

Lastly, title the graph. The title should be short and clear. In this case, title the graph "Television Sales."

The box-and-whisker plot should look like the one below.

#### Example 3

The box plot below shows the weight (in pounds) of black bears. Identify the minimum, maximum, median, first quartile, and third quartile of the data set.

First, find the minimum value of the data set. To do this look at the end of the left whisker. The left whisker ends at 127, so this is the minimum.

Next, find the maximum value of the data set. To do this look at the end of the right whisker. The right whisker ends at 201, so this is the maximum.

Next, find the first quartile. To do this look at the left side of the box. The left side of the box is at 129, so this is the first quartile.

Next, find the median. To do this look at the vertical line inside the box. This is at 163, so this is the median.

Next, find the third quartile. To do this look at the right side of the box. The right side of the box is at 196, so this is the third quartile.

The answer is the minimum is 127 pounds, the maximum is 201 pounds, the median is 163 pounds, the first quartile is 129 pounds, and the third quartile is 196 pounds.

#### Example 4

The values in the data set below represent the number of dogs that returned from a sled race. Create a box-and-whisker plot of the data.

11, 11, 12, 10, 9, 10, 13, 7, 9, 7

First, order the data from least to greatest.

7, 7, 9, 9, 10, 10, 11, 11, 12, 13

Next, find the median, which is the middle number. There are ten terms, so the median is between the fifth and sixth terms, 10 and 10. Since these terms are the same, the median is 10. The median will not be used in calculating the first or third quartiles.

Next, find the first quartile, which is the middle number of the lower half of the data set. Since 10 is not included, there are four terms in the lower half of the set, 7, 7, 9, and 9. The median of this set is between the second and third terms, 7 and 9. To find the median we add the two terms together and divide by 2.

7+9=16\begin{align*}7+9=16\end{align*}

16÷2=8\begin{align*}16÷2=8\end{align*}

So, the first quartile is 8. This will be the left side of the box.

Next, find the third quartile, which is the middle number of the upper half of the data set. Since 10 is not included, there are four terms in the upper half of the set, 11, 11, 12, and 13. The median of this set is between the second and third terms, 11 and 12. To find the median we add the two terms together and divide by 2.

11+12=23\begin{align*}11+12=23\end{align*}

23÷2=11.5\begin{align*}23÷2=11.5\end{align*}

So, the third quartile is 11.5. This will be the right side of the box.

Next, find the minimum and maximum of the data set. These will be the ends of the whiskers. The minimum is the smallest term in the set. In this case, it's 7. The maximum is the largest term in the set. In this case, it's 13.

Next, draw the box-and-whisker plot. Begin by drawing a number line that begins at the minimum, 7, and extends to the maximum, 13. In this case, the number line can begin at 7 and go to 13. Then draw the box-and-whisker plot above the number line.

Lastly, title the graph. The title should be short and clear. In this case, title the graph "Sled Dogs."

The box-and-whisker plot should look like the one below.

#### Example 5

The box-and-whisker plot below was shows 5-K race times for some runners. Identify the minimum, maximum, median, first quartile, and third quartile of the data set.

First, find the minimum value of the data set. To do this look at the end of the left whisker. The left whisker ends at 12, so this is the minimum.

Next, find the maximum value of the data set. To do this look at the end of the right whisker. The right whisker ends at 26, so this is the maximum.

Next, find the first quartile. To do this look at the left side of the box. The left side of the box is at 14, so this is the first quartile.

Next, find the median. To do this look at the vertical line inside the box. This is at 17, so this is the median.

Next, find the third quartile. To do this look at the right side of the box. The right side of the box is at 21, so this is the third quartile.

The answer is the minimum is 12 minutes, the maximum is 26 minutes, the median is 17 minutes, the first quartile is 14 minutes, and the third quartile is 21 minutes.

### Review

Use each set of data to work with box-and-whisker plots.

12, 13, 15, 17, 21, 22, 24, 26, 28, 30, 31

1. What is the median of the set of data?
2. What is the median of the lower quartile?
3. What is the median of the upper quartile?
4. What is the lowest value whisker?
5. What is the highest value whisker?
6. Use the data to create a box-and-whisker plot.

26, 27, 29, 30, 32, 35, 41, 42, 44

1. What is the median of the set of data?
2. What is the median of the lower quartile?
3. What is the median of the upper quartile?
4. What is the lowest value whisker?
5. What is the highest value whisker?
6. Use the data to create a box-and-whisker plot.

100, 105, 107, 109.110, 120

1. What is the median of the data?
2. What is the median of the lower quartile?
3. What is the median of the upper quartile?
4. What is the lowest value whisker?
5. What is the highest value whisker?

To see the Review answers, open this PDF file and look for section 11.10.

### Notes/Highlights Having trouble? Report an issue.

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### Vocabulary Language: English

Extremes

The extremes are the maximum and minimum values in a data set.

five point summary

The numbers needed to construct a box-and-whisker plot are called the five-point-summary. The five points are the minimum, the lower median (Q1), the median, the upper median (Q3), and the maximum.

line of fit

A line of fit is a straight or continuously curved line representing the trend of changes in the comparison of two data sets (or one set of bivariate data).

Median

The median of a data set is the middle value of an organized data set.

observed data

Observed data are the values that result from computations performed on the input variable.

Outlier

In statistics, an outlier is a data value that is far from other data values.

Quartile

A quartile is each of four equal groups that a data set can be divided into.

skewed

As with the horizontal skewing of a histogram, stem plots with a obvious skew toward one end or the other tend to indicate an increased number of outliers either lesser than or greater than the mode.

statistical correlation

Statistical correlation is a representation of possible related changes in values between the two sets of data.

trends

Trends in data sets or samples are indicators found by reviewing the data from a general or overall standpoint

uniform

A uniform shaped histogram indicates data that is very consistent; the frequency of each class is very similar to that of the others.

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text:
Get Systems - Maple Help

Units

 GetSystems
 list all systems of units

 Calling Sequence GetSystems()

Description

 • The GetSystems() function returns an expression sequence of all systems of units.

Examples

 > $\mathrm{with}\left(\mathrm{Units}\right):$
 > $L≔\mathrm{GetSystems}\left(\right)$
 ${L}{≔}{\mathrm{Atomic}}{,}{\mathrm{CGS}}{,}{\mathrm{EMU}}{,}{\mathrm{ESU}}{,}{\mathrm{FPS}}{,}{\mathrm{MKS}}{,}{\mathrm{MTS}}{,}{\mathrm{SI}}$ (1)
 > $\mathbf{for}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}i\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}L\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{do}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}\mathrm{print}\left(\mathrm{system:},i,\mathrm{value:},\mathrm{convert}\left(32.23\mathrm{Unit}\left('\mathrm{newton}'\right),'\mathrm{system}',i\right)\right)\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathbf{end}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{do}$
 ${\mathrm{system:}}{,}{\mathrm{Atomic}}{,}{\mathrm{value:}}{,}{3.912013859}{×}{{10}}^{{8}}{}⟦\frac{{\mathrm{E0}}}{{\mathrm{a0}}}⟧$
 ${\mathrm{system:}}{,}{\mathrm{CGS}}{,}{\mathrm{value:}}{,}{3.22300000}{×}{{10}}^{{6}}{}⟦{\mathrm{dyn}}⟧$
 ${\mathrm{system:}}{,}{\mathrm{EMU}}{,}{\mathrm{value:}}{,}{3.22300000}{×}{{10}}^{{6}}{}⟦{\mathrm{dyn}}⟧$
 ${\mathrm{system:}}{,}{\mathrm{ESU}}{,}{\mathrm{value:}}{,}{3.22300000}{×}{{10}}^{{6}}{}⟦{\mathrm{dyn}}⟧$
 ${\mathrm{system:}}{,}{\mathrm{FPS}}{,}{\mathrm{value:}}{,}{233.1200364}{}⟦{\mathrm{poundal}}⟧$
 ${\mathrm{system:}}{,}{\mathrm{MKS}}{,}{\mathrm{value:}}{,}{32.23}{}⟦{N}⟧$
 ${\mathrm{system:}}{,}{\mathrm{MTS}}{,}{\mathrm{value:}}{,}{0.03223000000}{}⟦{\mathrm{sn}}⟧$
 ${\mathrm{system:}}{,}{\mathrm{SI}}{,}{\mathrm{value:}}{,}{32.23}{}⟦{N}⟧$ (2)


text:
# Data Storage - Logical Block Addressing (LBA)

Logical block addressing (LBA) is a common scheme used for specifying the location of blocks of data stored on computer storage devices, generally secondary storage systems such as hard disks.

LBA has replaced the CHS scheme in order to overcome certain of its limitations.

LBA provides a simple linear address space to the host. The host only need to provide the LBA address without knowing anything of the physical sector positions.

For a 32-bit address mode, the system can support up to 232 blocks of capacity which is about 2T bytes. In a 64-bit address mode, the system capacity support up to 264 blocks. Without 48-bit LBA support, drivers with only 32-bits of addressing will only be capable of addressing up to 137 GB on the hard drive.

The term LBA can mean either:

to which it refers.

Logical Block Addressing (LBA) is also known as the BIOS INT 13h extensions drive geometry translation.

## 3 - The LBA scheme

LBA is a particularly simple linear addressing scheme.

Blocks are located by an index, with the first block being LBA=0, the second LBA=1, and so on.

The LBA scheme replaces earlier schemes which exposed the physical details of the storage device to the software of the operating system.

Chief among these was the cylinder-head-sector (CHS) scheme, where blocks were addressed by means of a tuple which defined the cylinder, head, and sector at which they appeared on the hard disk.

CHS didn't map well to devices other than hard disks (such as tapes and networked storage), and was generally not used for them.

SCSI introduced LBA as an abstraction. While the drive controller still addresses data blocks by their CHS address, this information is generally not used by the SCSI device driver, the OS, filesystem code, or any applications (such as databases) that access the “raw” disk. System calls requiring block-level I/O pass LBA definitions to the storage device driver; for simple cases (where one volume maps to one physical drive), this LBA is then passed directly to the drive controller.

## 4 - Conversion between CHS to LBA

CHS addresses can be converted to LBA addresses using the following formula:

LBA = (( C x HPC ) + H ) x SPT + S - 1

where,

• C, H and S are the cylinder number, the head number, and the sector number
• LBA is the logical block address
• HPC is the number of heads per cylinder
• SPT is the number of sectors per track


text:
## Display Groups?

Questions about the LÖVE API, installing LÖVE and other support related questions go here.
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mastastealth
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### Display Groups?

Been a while since I've been here. Recently I've been working on my first Android game using Corona, but thinking about the future remembered about the awesome Love2D project in hopes of one day porting my game to PC's as well!

Now, taking a quick skim through the wiki, and my own code, I believe 70-80% of it can be practically used as-is or with a few function swaps (image loading, etc.) However there's one bit that I didn't see an alternative for and that was display groups. They seem to act almost like "layers" for the scene, and believe are just tables with some extra metadata stuff? They allow manipulation of various display objects at the same time. Is there something similar, planned, or possible to add in love?

thelinx
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### Re: Display Groups?

Take a look at SpriteBatches.

mastastealth
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### Re: Display Groups?

Hmm, close, though I can probably modify my code to use them properly.

However, group display seems to be a little more general. For example, I see in SpriteBatch, to add an image I also have to give coordinates. Although this might work in my case (with modifications), but I don't specifically need to add coordinates to simply "add it to the group". Also, in my specific case of a mancala-like game, each of the 12 pits serves as a group, while the objects I stick in them are the stones. So, when I play, a for loop empties out the current pit's group into the player's "hand" group (an object can't be in 2 display groups at once. So adding to the hand automatically removes from the pit) and then drops 1 stone (with updated position) into a new pit. Does Spritebatch have a single remove function? Or can sprites be in multiple batches at the same time (that way I can add all current stones to new hand, clear old pit, add stones to new pits, then clear hand)?

For memory's sake, I made each of the 48 stones (the amount in your typical mancala game) is an individual object that specifcally is moved around the board (until captured) instead of deleting old objects, and recreating them over and over; so I'd like this "unique instance" ability to remain if possible.

bartbes
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### Re: Display Groups?

A SpriteBatch is a collection of Quads to be applied to an image, it sound different than what you're using right now, though I suppose that could also be emulated with some clever tables.
To answer your question, you can't remove an entry from a SpriteBatch due to the way they work.

kikito
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### Re: Display Groups?

mastastealth wrote:Hmm, close, though I can probably modify my code to use them properly.

However, group display seems to be a little more general. For example, I see in SpriteBatch, to add an image I also have to give coordinates. Although this might work in my case (with modifications), but I don't specifically need to add coordinates to simply "add it to the group". Also, in my specific case of a mancala-like game, each of the 12 pits serves as a group, while the objects I stick in them are the stones. So, when I play, a for loop empties out the current pit's group into the player's "hand" group (an object can't be in 2 display groups at once. So adding to the hand automatically removes from the pit) and then drops 1 stone (with updated position) into a new pit. Does Spritebatch have a single remove function? Or can sprites be in multiple batches at the same time (that way I can add all current stones to new hand, clear old pit, add stones to new pits, then clear hand)?

For memory's sake, I made each of the 48 stones (the amount in your typical mancala game) is an individual object that specifcally is moved around the board (until captured) instead of deleting old objects, and recreating them over and over; so I'd like this "unique instance" ability to remain if possible.
What's the advantage of using that versus using plain tables?

Code: Select all

local pits = {}
for i=1, 12 do table.insert(pits, {}) end
...

-- insert into pit 12
table.insert(pits[12], ball)

-- get all the balls in pit 12
pits[12]

-- remove all the balls from pit 12
pits[12] = {}

When I write def I mean function.

mastastealth
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### Re: Display Groups?

In my case, I think I might of actually been able to use normal tables instead (now that I look it over). Actually, I was thinking that with that "MiddleClass" library I would just create the most-basic (i.e. these normal table functions with a different name) version of these "Display Group" tables, just for compatibility's sake. (Wouldn't have to go through all my code to fix to normal tables).

However, if I remember correctly, one can manipulate the display group to modify all the display objects at once, which I don't think you can do with just normal tables (without having to write up special functions).

For instance, let's say (using your example) that I want all the balls in the pits to become transparent because I opened a menu or something. I would be able to say

Code: Select all

pits.opacity = 0.5

or something like that and that's it. With the tables I assume I'd need a for loop to manually set each ball's opacity individually. Correct? Same thing for moving, I can update the pits.x and pits.y and have all the children object balls adjust relatively to that, instead of one by one. Lastly, like I mentioned, inserting into a display group vs a table let's me forget about having to clear the old table, because it's automatic.

Of course you can probably make these functions yourself for the table, which I believe is what I first stated with my idea above about using the MiddleClass library, but as far as I know, your standard Lua table couldn't do the 3 above mentioned examples right off-the-bat...?

thelinx
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### Re: Display Groups?

Code: Select all

containerTable = {}
setmetatable(containerTable, {
__newindex = function(tbl, key, val)
for _, child in pairs(tbl) do
child[key] = val
end
end
})

However, if the containerTable is a class created with Middleclass, doing this would overwrite that metatable.
Fortunately, Middleclass allows you to create metafunctions in your regular class definition, like so:

Code: Select all

function containerTable.__newindex(tbl, key, val)
...
end

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