Okay, today's topic is going to be about solving
trigonometric equations using factoring. Um,
this is from trigonometry or MAC 1114. My
name is Kurt Overhiser, and I'll be your host
as we do, I think, about three examples. Uh,
the examples I'm writing, that we are going
to take are taken from the textbook entitled
Trigonometry, fifth edition, by Larson and
Hostettler. The first one we're going to look
at is found on page 232, um, problem number
23. It states that 3 times the tangent cubed
of x is equal to the tangent of x. It would
be tempting, when you look at this equation,
to go ahead and divide both sides by tangent,
and that way, on the left hand side, you would
get a tangent squared, but, uh, it's not appropriate
to do that here, because you may, uh, lost
some solutions. So what we're going to do
is, instead of dividing, we're going to be
subtracting the tangent of x from both sides.
And then, um, we're just going to go ahead
and, and factor this. So, if you can, if,
if you look to the, uh, side right here, where
my little arrow is pointing to, if you can
factor this, 3 y cubed minus 3 equals 0, from,
you know, just a regular beginning algebra
class, then you can factor this trigonometric
equation. It's basically the same thing. So
let's go ahead and factor the 3 y cubed minus
y equals 0. And we'll factor it out, we'll
factor out a greatest common factor of y,
like that. And then we can set each factor
equal to 0 using the zero-product property.
And so we end up with that. So we're going
to basically go backwards and, uh, use this
idea that we have, that we just worked here
on the side of the margin. And we're going
to, uh, factor out the greatest common factor;
instead of being y, it's going to be, we're
going to factor out the tangent of x. And
we're left with 3 times tangent squared of
x minus 1 equals 0. And then we're going to,
uh, set each factor equal to 0 because of
the zero-product property. And the tangent
of x is 0, uh, whenever the sine is 0, and
that happens at multiples of pi. So, uh, if
we let, uh, k be an integer, and then x equals,
uh, k times pi, um, and then for the other
one, we'll get tangent squared x equaling
1/3. And then if you, uh, square root both
sides, you can get tangent of x equals...
so we get, uh, tangent of x is equal to 1
divided by the square root of 3. And if you
know that that happens to be a special angle
of 30 degrees, so that would be pi over 6
radians, so x equals pi over 6, and then we're
going, since the period of the tangent function
is pi, we have to remember to add multiples
of pi to our, our answer. But we're still
not done, because when we, when we take this,
uh, when we extracted the root, when we took
the square root of both sides, we have to
remember to put a plus or minus in there,
and so the tangent of x is negative in quadrant
2, and that, uh, corresponding reference angle
in quadrant 2 would be 5 pi over 6, and then
we add to that multiples of pi to it. Again,
k is an integer, okay? So our, these are our
three solutions, and that is the end of example
1. Our next example is going to be, um, on,
taken from page 232, uh, number, homework
problem number 24. And this problem states
that you're supposed to solve this equation.
You're supposed to solve 2 times the sine
squared of x equals 2 plus cosine of x, and
we're going to solve it on the closed interval
from 0 to 2 pi. So what we'll do is, we have
to make use of a Pythagorean identity we're
going to be using; it's the most famous identity
of all. It says that sine squared x plus cosine
squared x equals 1. And if you solve for sine
squared x, you get 1 minus cosine squared
x. And this is the identity right here that
we're going to be using in this problem. So
the way we're going to do this is we're going
to, um, look at the sine squared that you
see right here, and we're going to rewrite
it, um, with that identity that we showed
you in the margin. So that gets rewritten
as 2 times 1 minus cosine squared x equaling
2 plus cosine of x. And then if you distribute
the 2, you get this. And then, what I'm going
to do next is I'm going to, uh, rearrange
some terms. Notice that there's a 2 on both
sides of the equation that you can, um, uh,
subtract 2 from both sides of the equation,
and they'll end up canceling from both sides.
And then what I'll do is I'll add the 2 cosine
squared x to both sides, so I'll end up with
2 cosine squared x 
plus cosine of x equals 0. And then, we're
going to, uh, do like before; we're going
to look over at the side over here. And if
you can factor this beginning algebra problem,
you can finish this trigonometric, uh, equation.
So the corresponding, uh, equation would be
2 y squared plus y is equal to 0. We're, in
essence, using a substitution where we are
letting y equal the cosine of x. So you factor
out the greatest common factor, which is y.
And then you set each factor equal to 0 using
the zero-product property. Okay? So we're
going to take, we're going to go back to our
original equation and, and, uh, factor it
in the same way that we just did in our margin,
except this time, instead of factoring out
a y, we're going to factor out a cosine of
x. And then, using the zero-product property,
we can get cosine of x equals 0, or 2 times
the cosine of x plus 1 equals 0, which also,
if you solve for cosine of x, you get cosine
of x equals negative 1/2. Okay, so we're almost
done here. Notice that we're working on the
unit circle from 0 to 2 pi, and we know that
the, uh, cosine of x is 0 in two places along
the, or around the unit circle. So here, for
cosine of x is 0, we get two answers. We get,
uh, if you remember your unit circle, pi halves
would be one place where the cosine is 0,
and then the second place would be 3 pi over
2. And then for the second factor, the cosine
of x equals negative 1/2, the cosine is negative
in quadrants 2 and 3, and so we're going to
get two answers for that. We're going to get
x is equal to 2 pi divided by 3, uh, because
the, the reference angle here is going to
be 60 degrees. So, and, uh, cosine's negative
in quadrant 2, so that would be 2 pi over
3. And then cosine is also negative in, in
quadrant 3, and that would give you, uh, 4
pi over 3. So these are the solutions to our
problem. Get four solutions: pi halves, 3
pi over 2, 2 pi over 3, and 4 pi over 3 for
this example. Okay, so our next example is
going to be taken, taken from page 228, example
number 6. And we're supposed to solve this
problem: 2 sine squared of x plus 3 cosine
of x minus 3 equals 0. Again, we're going
to use the Pythagorean identity that the sine
squared of x equals 1 minus cosine squared
x. So we use that right off the bat, and we
put 2 sine squared, replace it with 1 minus
cosine squared of x, plus 3 cosine of x minus
3 equals 0. And then we, if we distribute
the 2, we get this. And then what we're going
to do is just kind of, uh, rearrange some
of the terms around. And when I rearrange
the terms, I'm going to add 2 cosine squared
x to both sides as a first step, and then
a second step, I'm going to subtract, uh,
3 cosine of x from both sides, and then you'll
notice that 2 minus 3 is negative 1, and so
I'm going to add 1 to both sides as a third
step. So doing all three, all these three
steps in one, we get this: 2 cosine squared
x minus 3 cosine of x plus 1 equals 0. Okay,
and then for here, this happens to be a quadratic
in cosine of x, so if you look at our beginning
algebra example, uh, here's a good problem
that's associated with this problem, will
be to factor 2 y squared minus 3 y plus 1
equals 0. So to factor this, you put down
two sets of parentheses, and then you look
at the leading term 2 y squared, and the only
way to factor that would be 2 y and a y. You
look at the constant term plus 1, and that
means that they both are positive or both
negative, but because the negative's in the
middle, they both have to be negative, and
the only way to factor 1 is 1 and 1. So those
are the factors that we're going to be using.
Okay? And, uh, using this idea, we're going
to go back to the original problem, and, if
we think of y as the cosine of x as a, as
the basis for our substitution, we're going
to factor this into 2 times the cosine of
x minus 1 multiplied by the cosine of x minus
1 equals 0. And then we use the zero-product
property, and we do 2 times the cosine of
x minus 1 equals 0, or cosine of x minus 1
equals 0. That tells us that the cosine of
x is either equal to 1/2, or cosine of x equals
1, positive 1. Okay? And for the first factor,
where the cosine of x equals 1/2, uh, that's
positive, and we know that the cosine is positive
in quadrants 1 and 4. And in quadrant 1, we
would get pi thirds, and that's in quadrant
1. And then in quadrant 4, if you remember
your unit circle, we would get 5 pi over 3.
And then for the, the second factor, where
the cosine of x is equal to 1, um, that, we,
we can solve that easily, x has to be 0 right
there. Okay? Now in this problem, unlike the
previous problem, there, we're not restricting
ourselves to any finite interval, so, uh,
we have to solve this along the entire real
number line. So in order to do that, we have
to take advantage of the periodicity of the
cosine function. And its period is 2 pi, so
we have to take each of these answers that
we found in going around the unit circle once,
and then we have to add multiples of 2 pi.
So we will add 2 k pi, plus 2 k pi here, and
then plus 2 k pi here. K can be any integer,
um, both positive or negative or 0, and so
these would be our final results right here.
Okay? So that, um, we just went through, uh,
three problems where you did some factoring,
um, we did, first two problems, we factored
out a greatest common factor, we also had
to use, uh, some Pythagorean identities, and
the last one we had to solve, uh, using, by
factoring, uh, quadratic trinomial. So, hope
this helped, and good luck in your studies.
Okay? Bye-bye.
