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!

Hint:

Use $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\sin^2x=1-\cos^2x$

Update:
$$\cos x=-\frac{1}{2},1$$
$$\cos x=-\dfrac12=\cos\left(\frac{2\pi}3\right)\implies x=2k\pi\pm\dfrac{2\pi}3$$
$$\cos x=1\implies x=2k\pi$$
