;  Use the double angle identity for cosine.

;  Move everything to the left side of the equation.

;  This is a quadratic-like expression that cannot be factored.  We must use the quadratic formula.  It may be helpful to see this if you replace with , so it becomes:

Recall the quadratic formula 

plug in .

We now have

;  Separate this into two equations and take the inverse sine.

or 

The first equation gives us .  Using the unit circle as we did in previous problems, we can find a second answer from this which is .  The second equation will not give us a solution.

When the quadratic formula is applied to the function, it yields

So those are the zeros for sine, but sine has a minimum of -1, so -2 is out. For -1/2, sine achieves that twice in a cycle, at π+π/6 and 2π-π/6. So while -π/6 is true, it is not correct since it is not in the given interval.

Therefore on the given interval the zeros are:

Since can be written as:

. We can't have .

Therefore . This means that where k is an integer.

since . We have x=0 is the only number that satisfies this property.

Rearrange the problem,

Over the interval 0 to 360 degrees, cosx = 1/2 at 60 degrees and 300 degrees. 

First, get the equation in terms of one trig function. We can do this by substituting in using the Pythagorean Identity for .

Then we have .

Bring all the terms to one side to find that .

We can factor this quadratic to .

This means that .

The only angle value for which this is true is . 

First, re-write the equation so that it is equal to zero:

Now we can use the quadratic formula to solve for x. In this case, the coefficients a, b, and c are a=1, b=2, and c=-3:

simplify

This gives two potential answers:

and

Sine must be between -1 and 1, so there are no values of x that would give a sine of -3. The only solution that works is . The only angle measure that has a sine of 1 is .

Use the quadratic formula to solve for x. In this case, the coefficients a, b, and c are a=4, b=1, and c=-1:

simplify

the square root of 17 is about 4.123. This gives two potential answers:

. We can solve for x by evaluating both and . The first gives an answer of . Add this to 360 to get that as a positive angle measure, . If this has a sine of -0.64, so does its reflection over the y-axis, which is .

The second gives an answer of . If that has a sine of 0.39, then so does its reflection over the y-axis, which is .

There are multiple solution paths. We could subtract 1 from both sides and use the quadratic formula with and . Or we could solve using inverse opperations:

divide both sides by 2

take the square root of both sides

The unit circle tells us that potential solutions for are .

To get our final solution set, divide each by 3, giving:

.

 This problem has multiple solution paths, including subtracting 5 from both sides and using the quadratic formula with . We can also solve using inverse opperations:

subtract 2 from both sides

divide both sides by 4

take the square root of both sides

If the sine of an angle is , that angle must be one of . Since the angle is , we can get theta by subtracting :

To solve, use the quadratic formula with and where x would normally be:

This gives us two potential answers:

since this number is greater than 1, it is outside of the domain for cosine and won't give us any solutions.

Consulting the unit circle, the cosine is when