There’s not much to do with this one.  Just divide both sides by 2 and then 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.  Go back to my introduction in the Trig Function Evaluation 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.

This problem is almost identical to the previous except now I want all the solutions that fall in the interval .  So we will start out with the list of all possible solutions from the previous problem.

Then start picking values of n until we get all possible solutions in the interval.

First notice that since both the angles are positive adding on any multiples of  (n positive) will get us bigger than  and hence out of the interval.  So, all positive values of n are immediately out.  Let’s take a look at the negatives values of n.

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, the solutions are : .

This one is very similar to Problem 1, although there is a very important difference.  We’ll start this problem in exactly the same way as we did in Problem 1.

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?  A quick way of doing this is to, for a second, ignore the “-” in the problem and solve  in the first quadrant only.  Doing this give a solution of . Now, again using some geometry, this tells us that 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 looking for positive angles between 0 and .

Now we come to the very important difference between this problem and Problem 1.  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 I also divided the  by 5 as well!  This is important!  If you 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!

This problem is almost identical to the previous problem except this time we have an argument of  instead of .  However, most of the problem is identical.  In this case the solutions we get will be