Obtaining an exact solution to $\cos^{2}(x)=2\sin(2x)(\cos(x)-1)$ The title says the most of it. I need some way to generate an exact solution (if possible) to $$\cos^{2}(x)=2\sin(2x)(\cos(x)-1)$$ other than just graphing it and seeing its intersections on the x axis.
What I have originally tried so far is to make everything in terms of cos(x) or sin(x). For example, with u=2x, I get that
$$cos^{2}(x)=4sin(x)cos^{2}(x)-4sin(x)cos(x) \to cos^{2}(\frac{u}{2})=4sin(\frac{u}{2})cos^{2}(\frac{u}{2})-4sin(\frac{u}{2})cos(\frac{u}{2})$$
At this point, I can use the sine/cosine half angle identities to write everything in terms of cos(u), which would seem helpful to me
$$\frac{1+cos(u)}{2}=\sqrt{\frac{1-cos(u)}{8}}(\frac{1+cos(u)}{2})-\sqrt{\frac{1-cos^{2}(u)}{4}} \to$$
$$\frac{1+cos(u)}{2}=\frac{1}{4}\sqrt{\frac{1-cos(u)}{2}}(1+cos(u))-\frac{sin(u)}{2}$$
However at this point it seems like I am making the equation even worse... any solutions or ideas would be greatly appreciated!
The other thing I noticed is that I can divide through by a factor of cos(x) to get that 
$$cos(x)=2sin(x)(cos(x)-1) \to 2sin(x)cos(x)-2sin(x)-cos(x)+1=1 \to(2sin(x)-1)(cos(x)-1)=1$$
But I am still unsure as to how exactly that equation could be solved with exact solutions as well...
 A: As @user170231 stated above, we have 
$$\cos^{2}(x)-2\sin(x)\cos(x)(\cos(x)-1)=0\Rightarrow \cos(x)(\cos(x)-2\sin(x)(\cos(x)-1))=0\\ \Rightarrow \begin{cases}\cos(x)=0 \\ \cos(x)-2\sin(x)(\cos(x)-1)=0\end{cases}$$
The solution of $\cos(x)-2\sin(x)(\cos(x)-1)=0$ can be obtained by half angle tangent substitution $\cos(x)=\frac{1-\tan^2 (x/2)}{1+\tan^2 (x/2)}, \sin(x)=\frac{2\tan (x/2)}{1+\tan^2 (x/2)}$, therefore
$$\frac{1-\tan^2 (x/2)}{1+\tan^2 (x/2)}-2\frac{2\tan (x/2)}{1+\tan^2 (x/2)}\left( \frac{1-\tan^2 (x/2)}{1+\tan^2 (x/2)}-1 \right)=0 \\ \Rightarrow 1-\tan^2 \left(\frac{x}{2}\right)+8\tan^3 \left(\frac{x}{2}\right)=0$$
Thus if we consider $z=\tan \left(\frac{x}{2}\right)$ we have to find the roots of $8z^3-z^2+1=0$. This equation has only one real root (using some CAS) $z_1=\frac{1}{24} \left(1 - \frac{1}{\sqrt[3]{863 - 24 \sqrt{1293}}} - \sqrt[3]{863 - 24 \sqrt{1293}}\right)\approx -0.4616$.
A: Factorize as $\cos^{2}(x)=2\sin(2x)(\cos(x)-1)$ as
$$\cos x (1-4\sin x (\cos x-1) ) = 0$$
The factor $\cos x = 0 $ yields the solutions $x = \frac\pi2+n\pi$. For the second factor, use $\sin x = \frac{2t}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$, where $t = \tan\frac x2$, to get,
$$t^4-16t^3-1=0$$
which has two real roots. Unfortunately, there is no exact analytical expressions available. However, given the large coefficient 16, very good approximations can be obtained with $t\approx 16$ and $16t^3 \approx 1$, which leads to 
$$\tan\frac x2 = 16, \>\>\>\>\>\tan\frac x2 =-\frac1{\sqrt[3]{16}}$$
Thus, the full solutions are 
$$x = \frac\pi2+n\pi, \>\>\> 2\tan^{-1}16+2\pi n, \>\>\> -2\tan^{-1}\frac1{\sqrt[3]{16}}+2\pi n$$
Compared to the respective exact values $3.017+2\pi n$ and $-0.750+2\pi n$, the approximations $2\tan^{-1}16=3.017$ and $-2\tan^{-1}\frac1{\sqrt[3]{16}}=-0.755$ are pretty accurate.
