Example 1  Solve .

Solution

There’s really not a whole lot to do in solving this kind of trig equation.  All we need to do is divide both sides by 2 and the go to the unit circle.

So, we are looking for all the values of t for which cosine will have the value of .  So, let’s take a look at the following unit circle. 

From quick inspection we can see that  is a solution.  However, as I have shown on the unit circle there is another angle which will also be a solution.  We need to determine what this angle is.  When we look for these angles we typically want positive angles that lie between 0 and .  This angle will not be the only possibility of course, but by convention we typically look for angles that meet these conditions. 

To find this angle for this problem all we need to do is use a little geometry.  The angle in the first quadrant makes an angle of  with the positive x-axis, then so must the angle in the fourth quadrant.  So we could use , but again, it’s more common to use positive angles so, we’ll use .

We aren’t done with this problem.  As the discussion about finding the second angle has shown there are many ways to write any given angle on the unit circle.  Sometimes it will be  that we want for the solution and sometimes we will want both (or neither) of the listed angles.  Therefore, since there isn’t anything in this problem (contrast this with the next problem) to tell us which is the correct solution we will need to list ALL possible solutions. 

This is very easy to do.  Recall from the previous section and you’ll see there that I used

to represent all the possible angles that can end at the same location on the unit circle, i.e. angles that end at .  Remember that all this says is that we start at  then rotate around in the counter-clockwise direction (n is positive) or clockwise direction (n is negative) for n complete rotations.  The same thing can be done for the second solution.

So, all together the complete solution to this problem is

As a final thought, notice that we can get  by using  in the second solution.

Example 2  Solve  on .

Solution

In a calculus class we are often more interested in only the solutions to a trig equation that fall in a certain interval.  The first step in this kind of problem is to first find all possible solutions.  We did this in the first example.

Now, to find the solutions in the interval all we need to do is start picking values of n, plugging them in and getting the solutions that will fall into the interval that we’ve been given.

n=0.

Now, notice that if we take any positive value of n we will be adding on positive multiples of 2π onto a positive quantity and this will take us past the upper bound of our interval and so we don’t need to take any positive value of n.

However, just because we aren’t going to take any positive value of n doesn’t mean that we shouldn’t also look at negative values of n.

n=-1.

These are both greater than  and so are solutions, but if we subtract another  off (i.e use  ) we will once again be outside of the interval so we’ve found all the possible solutions that lie inside the interval .

So, the solutions are : .

Example 3  Solve  on  

Solution

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 solutions.

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 solutions!

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 we’ve got.

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 given interval.