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.
 A: 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.
A: We go some distance to a solution.  Might as well get rid of the silly extra $2$'s, and solve
$$\sqrt{1-\cos x}+\sqrt{5-3\cos x}=\sqrt{5-3\cos 2x}=\sqrt{8-6\cos^2 x}.$$
From now on write $w$ instead of $\cos x$. So we are solving
$$\sqrt{1-w}+\sqrt{5-3w}=\sqrt{8-6w^2}.$$
Square both sides, simplify a bit. We get 
$$\sqrt{1-w}\sqrt{5-3w}=-(3w^2-2w-1)=(1-w)(3w+1).$$
Note that each side has a $1-w$. Cancel, remembering the root $w=1$.  Then square both sides again. We get
$$5-3w=(1-w)(3w+1)^2.$$
A cubic! But not a terrible cubic, it is $9w^3-3w^2-8w+4=0$, which happens to have $-1$ as a root. So divide by $w+1$, we get a quadratic.  Solve, and throw
away anything that has snuck its way in as a result of the squaring process.
Remark: Things could have turned ugly. There was the happy "accident" that allowed us to partially get rid of a $1-w$ factor. And then there was the other happy accident of an obvious root $w=-1$.  A small perturbation of the numbers could make things difficult.
