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?

Notice that $\begin{cases}\cos(x)+\cos(y)+\cos(z)=0\\\sin(x)+\sin(y)+\sin(z)=0\end{cases}\iff \underbrace{e^{ix}}_a+\underbrace{e^{iy}}_b+\underbrace{e^{iz}}_c=0$
We have $a+b+c=0\implies \bar a+\bar b+\bar c=0$
But since $|a|=|b|=|c|=1$ they verify $\bar a=\frac 1a$ and so on.
Therefore we get $$\frac 1a+\frac 1b+\frac 1c=0\implies \dfrac{ab+bc+ca}{abc}=0$$
Note that $|abc|=1$ so the numerator should be zero.
But then $$\overbrace{(a+b+c)^2}^{=0}=\overbrace{(a^2+b^2+c^2)}^{e^{i2x}+e^{i2y}+e^{i2z}}+2\overbrace{(ab+bc+ca)}^{=0}$$
And your sum is simply $0$ too.
