We're looking for something, and that something is (something).
But first we need to isolate the function.
Looking at the unit circle, we see that the angles with y-coordinates of positive are and .
Those are the angles we want to equal (something), or 2t in this case.
These are all of the angles. All of them.
Example 2
Solve (cos (3t))2 + 2(cos (3t)) + 1 = 0 on the interval [0, π].
Well, this looks a bit different.
What happens if we replace cos (3t) with u?
u2 + 2u + 1 = 0
Oh, we can factor this!
(u + 1)(u + 1) = 0
Or, perhaps we should say:
(cos (3t) + 1)(cos (3t) + 1) = 0
If either part of this equation equals 0, then the whole thing will.
cos (3t) + 1 = 0
cos (3t) = -1
We know how to work with this.
According to the unit circle, we need π to get -1 for cosine.
3t = π + 2πn
Plugging in values of n, we see that n = 0 and 1 give valid results on the interval.
Example 3
Solve .
Cotangent equals . That means that . For this problem:
Checking the unit circle, we see that .
So does .
Notice that are π away from each other, so we could have consolidated our answer a bit. We don't have to, though; plug in a few values for n and be amazed as you get all of the same answers.
.
are 
and 
.
.
. That means that 
. For this problem:
.
.
are π away from each other, so we could have consolidated our answer a bit. We don't have to, though; plug in a few values for n and be amazed as you get all of the same answers.