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?

Although this is same to answer of @schrodingersCat　Let $x\geq y\geq0\geq z$ 
$$\cos(x+y)^2=1-\sin(x+y)^2$$ 
$$⇔1-(\sin{x}+\sin{y})^2=(\cos{x}+\cos{y})^2$$
$$⇔1+2(\cos{x}\cos{y}+\sin{x}\sin{y})=0$$
$$⇔\cos(x-y)=-1/2 $$
$$x-y=2\pi/3$$
and
$$\sin{x}+\sin{y}=-\sin{z}$$
$$⇔\sin(y+2\pi/3)+\sin{y}=\sqrt3/2$$ 
$$⇔(1/2)\sin{y}+\sqrt3/2\cos{y}=\sqrt3/2$$
$$⇔\sin(y+\pi/3)=\sqrt3/2$$ 
then we get $$x=2\pi/3 ,y=0,z=-2\pi/3$$, if angle(x,y,z) are points of unit circle, they consist of the regular triangle. The answer of $x^2+y^2+z^2$  is unique and,
$$\displaystyle x^2+y^2+z^2=\frac{8\pi^2}9 $$
