Solving $2\cos^2 x-2\sin^2 x-2\cos x=0$ $$f(x) = 2\cos^2 x-2\sin^2 x-2\cos x$$
Need values of x that which make $f(x) = 0$
Tried $a^2-b^2 = (a+b)(a-b)$ with no luck
Really just need a hint that could bring me in the right direction
Thanks
EDIT: Solution thanks to everyones help! :D 
$$f(x) = 2\cos^2 x-2\sin^2 x-2\cos x$$
$$0 = 2\cos^2 x-2 + 2\cos^2 x-2\cos x$$
$$0 = 4\cos^2 x-2\cos x - 2$$
$$\cos x = 2\pm \sqrt {-2^2-4(4)(-2)\over8}$$
$$\cos x =  {2 \pm 6\over 8}$$
$$\cos x = {-1\over2}, 1$$
$$x = 2n\pi \pm{2\pi\over3}$$
Thank you!

We have $$\cos x=\cos^2x-\sin^2x=\cos2x$$
$$\implies2x=2m\pi\pm x$$ where $m$ is any integer
Taking the '-' sign, $3x=2m\pi\iff x=\dfrac{2m\pi}3\  \ \  \ (1)$
Taking the '+' sign, $x=2m\pi $ which is a proper subset of $(1)$
