# Homework Help: A problem in finding the General Solution of a Trigonometric Equation

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1. May 18, 2017

### Wrichik Basu

1. The problem statement, all variables and given/known data:

Find the general solution of the Trigonometric equation $$\sin {3x}+\sin {x}=\cos {6x}+\cos {4x}$$

Answers given are: $(2n+1)\frac {\pi}{2}$, $(4n+1)\frac {\pi}{14}$ and $(4n-1)\frac {\pi}{6}$.

2. Relevant equations:

Equations that may be used:

3. The attempt at a solution:

The answer from Case 1 is correct, but I can't find my mistake in the answers from the two sub-cases of case 2.

2. May 18, 2017

Your mistake is in the trigonometric identity $cosA-cosB=-2sin((A+B)/2)sin((A-B)/2)$. You didn't divide the terms (A+B and A-B) by 2 in both cases.

3. May 18, 2017

### Wrichik Basu

Got it. Thanks a lot.

4. May 18, 2017

@Wrichik Basu This is an extra detail, but it may interest you that I think the solution $x=(2n+1) \frac{\pi}{2}$ for all integers $n$ is actually all included in the other two solutions. The reason is that $x=(2n+1) \frac{\pi}{2}$ is also always a solution of $cos(5x)=sin(2x)$. (A complete expansion of $cos(5x)$ and $sin(2x)$will generate a $cos(x)$ factor on both sides of the equation.) $\\$ You can write $x= (2k+1) \frac{\pi}{2}=(4n+1) \frac{\pi}{14}$ and if $k$ is odd, for any $k$ you can find an integer $n$. You can also write $x=(2k+1) \frac{\pi}{2}=(4m-1) \frac{\pi}{6}$ and if $k$ is even, for any $k$ you can find an integer $m$. Thereby, the last two solutions completely overlap the $x=(2n+1) \frac{\pi}{2}$ solution. $\\$ Editing... The other two solutions are completely independent of each other=a little algebra shows there is no "x" that is the same in both of them.