Complex trinomial factoring $ 2 \cos x - 2 = \sin^2 x$ $2 \cos x -2 = \sin^2 x$
I have been trying to solve this equation for the interval $0 \le x \le 2\pi$ . 
I figured I should keep them as one, so I put
$2 \cos x -2 = 1 - \cos^{2} x$
however I don't really know how to proceed from there, and i'm pretty sure i'm not actually solving it correctly. I don't know what i'm doing wrong. 
 A: $$2 \cos x -2 = \sin^2x$$
$$\Rightarrow 2 \cos x -2 = \frac{1-\cos(2x)}{2}$$
$$\Rightarrow 4 \cos x -4 = 1-\cos(2x)$$
$$\Rightarrow 4 \cos x +\cos(2x) = 5$$
We then note that the $\cos x$ has a range of $[0,1]$, and thus our equation is only equal to $5$ when $\cos(2x) = 1$ and $\cos x = 1$. These both happen at interval of $2\pi$ (think of the unit circle here... only when it is $0$ degrees), and so our answer is $x = 2\pi n$, where $n$ is any integer. We have restricted our domain though from $0$ to $2\pi$, so the answer is $x=0,2\pi$
A: Hint: collect terms and treat the result as a quadratic equation where the variable is $\cos x$.
A: HINT:
$$2\cos(x)-2=\sin^2(x)\Longleftrightarrow$$
$$-2+2\cos(x)-\sin^2(x)=0\Longleftrightarrow$$
$$-3+2\cos(x)+\cos^2(x)=0\Longleftrightarrow$$
$$(\cos(x)-1)(3+\cos(x))=0\Longleftrightarrow$$
$$\cos(x)-1=0\Longleftrightarrow \space\space\vee\space\space 3+\cos(x)=0\Longleftrightarrow$$
$$\cos(x)=1\Longleftrightarrow \space\space\vee\space\space \cos(x)=-3\Longleftrightarrow$$
$$x=2\pi n_1 \space\space\vee\space\space x=\pm\cos^{-1}(-3)+2\pi n_2$$
With $n_1,n_2\in\mathbb{Z}$
A: HINT:
For $$ 2 \cos x - 2 = \sin^2 x$$
Note that $$\sin^2 x = {1-\cos^2x}$$
$$\implies2 \cos x - 2 = {1-\cos^2x}$$
Now solve the quadratic in disguise by letting $u=\cos x$
