Pre-Calculus: Solving Trigonometric Equations

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Trigonometric Identities
Solving Trigonometric Equations
Solving Trigonometric Equations Page [1 of 3]
Let’s now start taking a look at how we could solve very simply equations that involve trig functions. So now instead
of solving regular equations where we just have x’s and x2’s and so forth, let’s take a look at how to solve equations
that actually have trig functions in them. Let’s begin with ones to get us going. How about
cos 1
2
x = . Now, first of
all, let’s think about what that means and then see how we go about solving it. If I take a look at the cosine function, I
want to know what values of x make the cosine equal to
1
2
. So this is 1, here’s
1
2
. If I put that line there, the first
thing you see is that there are a lot of places, right? There’s a place over here, some x value that makes this equal to
1
2
, and there’s a place here, and there’s also a place way over here, too. The question is, how can I find all of those
solutions?
Well, let’s think about it. First of all, what you may want to think about is, well, what are those angles sort of roughly?
Well, one is going to be sort of small, between zero and
2
?
. So this is in fact a standard answer to a question, and
the question is, what standard angle has
cos 1
2
= ? Well, if you think about it, it’s 60! , also known as
3
?
. That’s one
of those standard ones that I hope that you may remember. So in fact, this answer right here, that must be at
3
?
. So
x could equal
3
?
.
But now wait a second. How do I get this answer here? Well, that’s something between
3
2
?
and 2? , and if you look
at the symmetry of this picture, you can actually see what I’m doing here. What I’m actually doing here is the
following. This is
3
?
here. So from this big peak out to the one I want is
3
?
. That means from this big peak
backwards out to here must be
3
?
, too, because this thing has sort of perfect symmetry. See how that length is the
same as that length? So to find this point, what I should do is take 2? and subtract
3
?
.
Let me show it to you in a different way. Maybe this way will make it a little bit easier to see. If we go back to the
angles themselves, just the angle part, what I see is the following. One answer is 60! , that’s
3
?
. But where else is
the cosine going to be positive? I want the cosine to equal positive
1
2
. All Students Take Calculus. So over here is
the other place where cosine is positive. So I have to have a reference angle of 60! or
3
?
, so I go down like this, and
now this is a new angle. And how do I find that new angle? Well, look what I do. I take 2? and subtract
3
?
. That’s
exactly what I said here: take 2? and subtract
3
?
. That angle corresponds to this angle right here. You can think of
Trigonometric Identities
Solving Trigonometric Equations
Solving Trigonometric Equations Page [2 of 3]
it this way if you want: All Students Take Calculus. Here the cosine is positive, so I must be down here. It’s 2
3
? ?? .
Well, the other way is to look at this picture and see that in the graph. In any case, what I see is that one answer is
this and the other answer is going to be 2
3
? ?? , and what is that? Let’s work that out.
2
3
? ?? , I have to get a common denominator so I multiply top and bottom here by 3, so I have
6
3 3
? ?? . 6? ? ? is
5? , over 3. So I see
5
3
?
. So there’s another answer. But in fact, there are a lot more answers, because there’s an
answer right here, for example. But notice that answer is the exact same sort of value as this if I just shifted it over
2? . See how if I just take that curve right there and shift it over, I get right to here. And there will be another answer
way over here when this bends down, but that’s the same thing as taking this answer and moving it over to here. So
I’m allowed to now add any multiple of the period, and the period is 2? , so 2? n I can add, and +2? n , where n is
any number—zero, ±1, ±2 , and so on. So in fact, there are infinitely many solutions; two sort of fundamental
solutions and then you can look at any sum. If you add on any multiple of 2? to these solutions, you’ll get other
solutions. You can see that graphically. So that’s the way I can solve sort of a straightforward thing like this.
Let’s try one more just for practice. Let’s take a look at
sin 2
2
x = ? . Let’s take a look at a picture first, a graph. I
want sin x to be
2
2
? , so negative, that’s down here somewhere. So where is that going to happen? Well, you
can see again that it happens a lot of places. It crosses it here, so at that x value right there. It also crosses here, so
at that x value. The two x values right here. And let’s see, over here there’s one, too, and so forth. So how do I find
those? Well, first I look at the sort of reference angle. I just look and ask, where does
sin 2
2
x = ? Well, that
happens at 45° or
4
?
. But now I see the sign is negative, so which quadrant should I be in? Well, this actually tells
me which quadrant I should be in. I should be in the ? to
3
2
?
, that’s third quadrant. And
3
2
?
to 2? , that’s the
fourth quadrant. And so visually the way you might want to think about this is as follows. If you think about the angle
itself, I’m going to want an angle down here, and this reference angle to be 45° or
4
?
. And the same thing here,
where this reference angle is
4
?
. So what are the angles themselves? Well, one angle looks like this and the other
angle goes way over to here.
How can I find those angles? Well, this angle—just look at this picture. This is going to be the straight angle, which is
? , plus
4
?
. So this is going to be
4
? +? , which if you multiply top and bottom here by 4, you’ll see
5
4
?
. So that’s
this angle right here,
5
4
?
. And how do I get that humongous angle? Well, one way of getting it is taking 2? and
subtracting
4
?
, going once around and then removing that reference angle. So if I do that, this angle here, that little
Trigonometric Identities
Solving Trigonometric Equations
Solving Trigonometric Equations Page [3 of 3]
angle is found, this. That huge angle, how would I find that? That huge angle would be just 2
4
? ?? . Well, 2? is
actually
8
4
?
, minus 1 would be
7
4
?
. So that huge angle is actually
7
4
?
. That’s the second solution we’ve found
here. That must happen at
7
4
?
.
But of course, there are a lot of solutions, and those other solutions can be found now by just adding multiples of 2? ,
adding multiples to periods. The answer would be
5
4
x = ? plus any multiple of 2? . And there’s another solution, a
whole bunch of other families of solutions,
7
4
?
plus any multiple of 2? . So there’s the answer, where again I remind
you, n can be anything of the form zero, ±1, ±2 , and so on.
So in fact, you can see when you’re solving a trigonometric function, there appear to be always infinitely many
solutions. And in general, there may be a lot of solutions. And the first thing to do is to sort of find the basic solution,
almost reference angle type things here, the core solutions, and then simply all the multiples that you can add to those
that still make it a solution.
Up next we’ll take a look at some slightly more exotic such equations and see if we can solve them using all the
methods we’ve talked about so far. Good luck.