Solve $ \sqrt{2-2\cos x}+\sqrt{10-6\cos x}=\sqrt{10-6 \cos 2x} $ $$ \sqrt{2-2\cos x}+\sqrt{10-6\cos x}=\sqrt{10-6 \cos 2x} $$
I tried squaring and/or using $1-\cos x=2\sin^2{\frac{x}2}$, but no luck.

If $t = \cos(x)$, we have $\sqrt{2-2t} + \sqrt{10-6t} = \sqrt{16-12 t^2}$.  Square both sides, isolate the term with square roots, square again, and factor.  The result should be equivalent to $(t+1)(t-1)(3t-2)^2=0$.  $t=-1$ does not work, but the other factors do.
