Today, we are going to look at trig equations
that require 
the quadratic formula 
or that can be solved by factoring.
Suppose we want to solve the following example
problem.
So we want to solve for x.
So we get the cosine of 2x is equal to the
sine of x, and you want the solution for all
real values of x.
Well, the first thing for us to do is reconcile
that we have a double-angle formula involving
cosine, and that's equaling the sine function,
so we're going to have to make a decision
about how to proceed here.
One of the benefits of having cosines and
sines is that there are so many identities
that relate to the two of them.
If we start with the double-angle formula
for cosine, that is equal to the cosine squared
x minus the sine squared x.
I always encourage my students to remember
where that equation is derived from.
If we start with the sum formula for the cosine
of two angles, cosine of x plus y is equal
to the cosine x cosine y minus the sine x
sine y.
If we let y equal x, then this problem becomes
the cosine, or this equation becomes the cosine
of x plus x equals the cosine x cosine x,
because y was replaced with x, minus the sine
x sine x, and that simplifies as the cosine
squared x minus the sine squared x.
So many of the identities, if you can remember
where they come from, you can re-derive them
in the event that you happen to forget this.
Next, let's use the Pythagorean triple, or
the Pythagorean identity that relates cosine
square and sine squared, namely, that sine
squared x plus cosine squared x equals 1.
Since this original equation involves solving
an equation set equal to the sine x, it would
be beneficial if we can rewrite this cosine
squared x minus sine squared x expression
strictly in terms of sines, and that can be
accomplished by taking, in this Pythagorean
identity, solving for cosine squared x, and
we get 1 minus sine squared x.
So that gets substituted in this double-angle
formula for the cosine of 2x.
Finally, we're able to re-derive the cosine
of 2x as 1 minus 2 sine squared x.
Again, if you can remember this, where it
comes from, it's a lot better than trying
to remember it, because as we move away from
this particular subject, memory fails, and
we may forget it.
We may remember it incorrectly, but we can
always derive it and see whether the equation
makes sense or not.
So this original example that we have up here
can be rewritten as the following equation,
solving for x for all real values of x: 1
minus 2 sine squared x equals sine x.
This is a quadratic in sine of x.
The way we solve quadratic equations are either
by factoring, and we can apply the 0 product
property, or by using the quadratic formula.
In any event, both require that we have this
equation set equal to 0.
So, if I take this equation and move everything
over to one side, I can get, as the following
quadratic equation: 2 sine squared x plus
sine x minus 1 is equal to 0.
This appears to be factorable.
What I think about is, say, if I were to let
u equal sine of x, then I can get an interesting
quadratic in u.
This is clearly solvable by applying the factoring
and then the 0 product property.
I would factor this quadratic as 2u and u
and a positive 1 and minus.
We'll verify that it produces a linear term,
linear coefficient term of 1.
This is -1u plus 2u, so this works.
With this substitution, realizing that u has
the same role as sine of x, we can factor
this quadratic in sine x as follows: 2 sine
x minus 1 multiplied by sine x plus 1 equals
0.
Therefore, the two solutions are sine x equals
positive 1/2, or sine x is equal to minus
1.
To solve problems involving integers, integral
values, 0, -1, 1, and, I guess, 0, 1, and
-1, those are the main values that I think
about, you can think about the unit circle.
I personally like to look at the sine graph
and just think about where this happens.
We know what happens here at 3pi over 4 radians,
3pi over 2 radians, rather.
If I were to draw one more cycle in this direction,
this would correspond to an x value of -pi
over 2 radians.
So solutions where the sine function's equal
to negative one are infinitely many, and I
can simply choose one, I can choose one that's
in the principle branch of the sine if I wanted
to, if I were interested in solving this by
applying the arcsine, or I can pick this value
here that's in, 3/4 of the way around a complete
revolution, so of course, in the context of
the unit circle, we're talking about the negative
y direction, the negative y axis, so that
would be 3pi/2 radians.
So it really doesn't matter which one you
choose.
I'm going to go with the -pi/2 radians, and
I'm going to say one set of solutions, which
would be indexed to integers, would be -pi/2
as well as every even multiple of pi and k
in this case would be an integer.
So we're going to use this symbol, it's just
the symbol of integers.
This obviates the need of having to include
a plus or minus sign, because integers include
positive integers, positive numbers, whole
numbers, zero, and negative natural numbers,
if you will.
So we don't need to put a plus or minus sign.
The next solution that we have to solve is
sine of x is equal to 1/2.
So, the sine is positive, we know that happens
in two quadrants.
So the two quadrants are the first and the
second.
So some values of x here could be in the first
quadrant, opposite over hypotenuse, we recognize
this as a 1-2-radical 3 right triangle, which
is our famous 30-60-90 right triangle.
Sine is also positive in the second quadrant.
This is opposite over hypotenuse, this would
be negative radical 3.
So these reference angles would be either
pi/6 radians, it's opposite the smallest side,
therefore it's the smallest angle, which corresponds
to 30 degrees.
And another set of solutions would begin—have
a reference angle of pi/6, which, an angle
in the second quadrant would be pi/6 short
of a complete pi, so it would be 5pi/6.
And again, we're talking about even multiples—integral
multiples of pi, where again, k is indexed
to the integers, and again, we're indexing
it to the integers.
If you want to index it to the whole numbers,
you can, and all you need to do is include
a plus or minus sign.
Thus we have our three series of solutions,
our three sets of solutions consisting of
three elements, all of them indexed to the
same parameter k, which are integers.
So the first set of solutions our x of k is
equal to -pi/2 plus 2kpi where k is an integer,
that corresponds to solutions to the form
sine of x is equal to -1.
And then we have solutions of the form x of
k is equal to pi/6 plus 2kpi, and 5pi/6 plus
2kpi, and those are solutions to the equation
sine of x is equal to 1/2.
I hope that was able to help you solving these
quadratics.
Thank you.
