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.

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I can give you $x=2k \pi$. –  Pedro Tamaroff Mar 8 '12 at 5:40
Is that the only solution, if so, can it be proven? –  Lazar Ljubenović Mar 8 '12 at 5:41
It isn't. That is the easy solution, since it will make $2-2\cos(x)$ zero and $10-6\cos(x)$ and $10-6\cos(2x)$ equal to four, so that you get $2=2$. –  Pedro Tamaroff Mar 8 '12 at 5:43
Squaring works fine. A little tedious. –  André Nicolas Mar 8 '12 at 5:51

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.
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.