Example 3 Solve
This problem is very similar to the other problems in this
section with a very important difference.
We’ll start this problem in exactly the same way. We first need to find all possible
So, we are looking for angles that will give out of the sine function. Let’s again go to our trusty unit circle.
Now, there are no angles in the first quadrant for which
sine has a value of . However, there are two angles in the lower
half of the unit circle for which sine will have a value of . So, what are these angles? We’ll notice ,
so the angle in the third quadrant will be below the negative x-axis or . Likewise, the angle in the fourth quadrant
will below the positive x-axis or . Remember that we’re typically looking for
positive angles between 0 and .
Now we come to the very important difference between this
problem and the previous problems in this section. The solution is NOT
This is not the set of solutions because we are NOT
looking for values of x for which ,
but instead we are looking for values of x
for which . Note the difference in the arguments of the
sine function! One is x and the other is . This makes all the difference in the world
in finding the solution! Therefore, the set of solutions is
Well, actually, that’s not quite the solution. We are looking for values of x so divide everything by 5 to get.
Notice that we also divided the by 5 as well! This is important! If we don’t do that you WILL miss solutions. For instance, take .
I’ll leave it to you to verify my work showing they are
solutions. However it makes the
point. If you didn’t divided the by 5 you would have missed these
Okay, now that we’ve gotten all possible solutions it’s
time to find the solutions on the given interval. We’ll do this as we did in the previous
problem. Pick values of n and get the solutions.
n = 0.
n = 1.
n = 2.
n = 3.
n = 4.
n = 5.
Okay, so we finally got past the right endpoint of our
interval so we don’t need any more positive n. Now let’s take a look
at the negative n and see what
n = 1
n = 2.
n = 3.
n = 4.
And we’re now past the left endpoint of the interval. Sometimes, there will be many solutions as
there were in this example. Putting
all of this together gives the following set of solutions that lie in the