First, using the trigonometric identity , re-write the expression so it is entirely in terms of :

 if we add  to both sides and re-order the terms, we can see that this is a quadratic equation in terms of sine:

 so we can use the quadratic formula to solve this way:

This gives us two potential answers:

this is outside of the range of sine, so it won't work.

Taking the inverse on a calculator yields , which we can add to 360 to give .

That's just one answer, however. To get the other answer, we need to know the other angle below the x-axis, so we can add.

We can factor this expression by grouping.

re-arrange the original expression so that the terms with sine are next to each other

factor our the

this factors to

Since we are finding the roots, we should set both factors equal to 0:

add 1 to both sides

subtract 1 from both sides

divide by 2

There are many ways to solve this problem, and here is one.

First, re-write the equation so that it is set equal to zero, since we're finding the roots, and so that it is all in one fraction. That is easy to do since both terms already have the common denominator of 3:

multiply both sides by 3

add the sine term to both sides

divide both sides by 2

multiply by 2

Factor the equation by grouping: the first 2 terms have in common, and the second 2 have 5 in common:

Since we are finding the roots, we can set each factor equal to 0:

subtract 5 from both sides

this is outside of the range for sine

add 1 to both sides

divide both sides by 2

take the square root of both sides

multiply by

That second one is greater than , so we want to figure out which angle between 0 and is coterminal with it. We can figure this out by subtracting , or in this case to get

Factor by grouping. The first 2 terms have in common.

We are finding the roots, so set each factor equal to 0:

subtract 1 from both sides

divide by 4 

add to both sides

divide both sides by 2

Factor the equation:

The roots occur when each factor equals 0:

Take the square root of both sides:

This gives us two roots, 

 and 

Solving for x:

 or ,

which evaluate to  and 

Now we check to make sure both answers work. Make sure your calculator is in radians mode!

Our answers are

if  is  

;  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: