Q:

Find roots of a trigonometric equation

I've been struggling on find the roots of this equation for a while.
$$
f(x)=\cos(2x) - 2\sin(x)\cos(x).
$$
I've already ended transforming all $\cos$ relation to $\sin$ ones, but I'm now stuck on:
$$
f(x)=1-2\sin^2(x) - \sin(2x).
$$
What should I do now?

A:

You have 

$$f(x)=\cos(2x)-2\sin(x)\cos(x).$$ 

The roots will satisfy $f(x)=0$, i.e. $$\cos(2x)-2\sin(x)\cos(x)=0$$ and so $\cos(2x)=2\sin(x)\cos(x)$. 
Now, $2\sin(x)\cos(x)=\sin(2x)$, so you have $$\cos(2x)=\sin(2x),$$ and from here, hopefully you can see what to do.

A:

Since
$$
f(x)=\cos(2x)-2\sin(x)\cos(x)=\cos(2x)-\sin(2x)=\cos(2x)-\cos(2x-\frac\pi2),
$$
we have
$$
f(x)=0\iff \cos(2x)=\cos(2x-\frac\pi2) \iff 2x=-2x+\frac\pi2+2k\pi, \ k \in \mathbb{Z},
$$
i.e.
$$
x=\frac\pi8+k\frac\pi2,\quad k \in \mathbb{Z}.
$$

