Find the sum of all possible values of $\cos 2x + \cos 2y + \cos 2z.$ Let $x$, $y$, and $z$ be real numbers such that $ \cos x + \cos y + \cos z = \sin x + \sin y + \sin z = 0 $. Find the sum of all possible values of $ \cos 2x + \cos 2y + \cos 2z $.
Here is what I have done so far
$$ \cos x + \cos y = -\cos z $$
$$ (\cos x + \cos y)^{2} = (-\cos z)^{2} $$
$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y = \cos^{2}z $$
likewise,
$$ \sin x + \sin y = -\sin z $$
$$ (\sin x + \sin y)^{2} = (-\sin z)^{2} $$
$$ \sin^{2} x + \sin^{2} y + 2 \sin x \sin y = \sin^{2}z $$
from this, you get
$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y + \sin^{2} x + \sin^{2} y + 2\sin x \sin y = \sin^{2}z + cos^{2}z $$
$$ 1 + 1 + 2(\cos x \cos y + \sin x \sin y) = 1 $$
$$ \cos x \cos y + \sin x \sin y = -\frac{1}{2} $$
$$ \cos(x-y) = -\frac{1}{2} $$
$$ x-y = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$
likewise,
$$ \cos (x-z) = -\frac{1}{2} $$
$$ x-z = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$
where do I go from here?

Continuing from your work, we have $y-z\equiv z-x\equiv x-y\equiv\frac23 r\pi\;\,(\!\!\!\!\mod2\pi)$, with $r\in\{1,2\}$ ($r$ is constant, since the three differences sum to zero, modulo $2\pi$). So $$\cos2x+\cos2y+\cos2z=\cos2x+\cos(2x-\tfrac43r\pi)+\cos(2x+\tfrac43r\pi)\qquad\qquad$$
$$=\cos2x+2\cos2x\cos\tfrac43r\pi$$
$$=0.\qquad\qquad\qquad\qquad\quad$$
