Find horizontal tangent line based on interval 
Find the values of $x$ on the interval $[−2\pi, 0]$ where the tangent line to the graph of $y = \sin(x)\cos(x)$ is horizontal.

I found a problem similar to this one, but I got lost when they magically put $\cos(\pi - x)$.
This is what I got: -(π/4),-(3π/4) Since the derivative is cos(2x), but it is still inccorect
 A: The derivative is cos(2x) we find the values of x when cos(2x) = 0 since the x is being multiplied by 2, this means there are twice as many answers for when cos(x) would equal 0. This means, using the unit circle, we can find the radians for which cos(x) = 0 and then take that radian and divide that by 2. Then just do an additional rotation, since it has twice as many answers, to get all four x values. Alternatively, you can use a graph to determine the answers by looking when y touches 0 
NOTE: Since the interval is in the negative, this means rotations are also negative. In other words, all answers should be negative. Although you can make every answer you get negative and get away with it, you must be careful if the interval isn't a full rotation (i.e -π) because (-π/2) is different from (π/2) despite their similar values
A: Hint: Just find the values of $x$ in that interval where $\frac{dy}{dx}=0$.
It may be easier if you rewrite $y=\sin x \cos x$ as $y=\frac12\sin 2x$ first.
A: You are asked for those points where $y'=0$. You have 
$$
y'=\cos^2x -\sin^2 x = 2\cos^2x-1. 
$$
So you are looking for those $x$ such that $2\cos^2x-1=0$, or
$$
\cos^2x=\frac12.
$$
So you want $\cos x=\frac1{\sqrt2}$, which happens at $x=\frac\pi4-2\pi=-\frac{7\pi}4$ and at $x=-\frac\pi4$. And also $\cos^2x=-\frac1{\sqrt2}$, which happens at $x=\frac{3\pi}4-2\pi=-\frac{5\pi}4$ and $x=\frac{5\pi}4-2\pi=-\frac{3\pi}4$. 
