<s>[INST] Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$ Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$.
My attempt:I found one answer $x=0,y=\frac{2\pi}{3},z=-\frac{2\pi}{3}$ which gives the answer $x^2+y^2+z^2=\frac{8{\pi}^2}{9}$.But I want a way to find the answer using equations.ANy hints?
 [/INST] You can try this method:
$$x+y+z=0 \implies x+y=-z\tag1$$
And $$\sin{x}+\sin{y}+\sin{z}=0$$
$$\implies \sin{x}+\sin{y}=-\sin{z}\tag2$$
And $$\cos{x}+\cos{y}+\cos{z}=0$$
$$\cos{x}+\cos{y}=-\cos{z}\tag3$$
So $$(2)^2+(3)^2 \implies 2+2\cos(x-y)=1$$
$$\implies \cos(x-y)=-\frac12\tag4$$
Similarly we get $$\cos(y-z)=-\frac12\tag5$$
$$\cos(z-x)=-\frac12\tag6$$
And from $(1)$, we get $\cos(x+y)=\cos z$
Can you proceed now?
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