Solve system $\cos^2x+\cos^2y+\cos^2z=1,$ $\cos x+\cos y+\cos z=1,$ $x+y+z=\pi$ I want to solve the system:

$$\cos^2(x)+\cos^2(y)+\cos^2(z)=1,$$
$$\cos(x)+\cos(y)+\cos(z)=1,$$
$$x+y+z=\pi.$$

I tried to prove that only one of cosines can be not a zero, but I just prove that one or three cosines can not be zero.
I get, that $$\cos(x)~\cos(y)+\cos(x)~\cos(z)+\cos(y)~\cos(z)=0$$
Also after substituting
$$\cos(x) = 1 - \cos(y)-\cos(z),$$
we get $$\cos^2(y) + \cos^2(z)=\cos(y)+\cos(z) - \cos(y)\cos(z)$$
 A: Since $z=\pi-x-y$, you have $\cos z=\cos(x+y)$; then
\begin{align}
\cos^2x&+\cos^2y+\cos^2z
\\[6px]
&=\cos^2x+\cos^2y+\cos^2x\cos^2y-2\cos x\cos y\sin x\sin y+\sin^2x\sin^2y
\\[6px]
&=\cos^2x+\cos^2y+\cos^2x\cos^2y-2\cos x\cos y\sin x\sin y
\\
&\qquad+1-\cos^2x-\cos^2y+\cos^2x\cos^2y\\[6px]
&=1+2\cos x\cos y(\cos x\cos y-\sin x\sin y)\\[6px]
&=1+2\cos x\cos y\cos z
\end{align}
which is an interesting identity of its own.
In your case, we get $\cos x\cos y\cos z=0$.
Assume $\cos x=0$; then $x=\pi/2+m\pi$ and $\cos y+\cos z=1$. By squaring we get also $\cos y\cos z=0$, so $y=\pi/2+n\pi$ or $z=\pi/2+n\pi$.
In the first case, $\cos z=1$, so $z=2k\pi$ and we also need
$$
\frac{\pi}{2}+m\pi+\frac{\pi}{2}+n\pi+2k\pi=\pi
$$
so $m+n+2k=0$. Similarly in the second case and for $\cos y=0$ or $\cos z=0$.
If you also have the limitation that the angles are those of a (possibly degenerate) triangle, so in $[0,\pi]$, you get that two of the angles are $\pi/2$ and the other one is $0$.
A: The hint.
Since $z=\pi-x-y$, we obtain
$$\cos{x}+\cos{y}-\cos(x+y)=1$$ or
$$2\cos\frac{x+y}{2}\cos\frac{x-y}{2}=2\cos^2\frac{x+y}{2}$$ or
$$\cos\frac{x+y}{2}\left(\cos\frac{x-y}{2}-\cos\frac{x+y}{2}\right)=0$$ or
$$\sin\frac{z}{2}\sin\frac{x}{2}\sin\frac{y}{2}=0.$$
I think the rest is smooth.
