Solve the system of equations: $\left\{\begin{array}{c}\sin y-\sin x=x-y \\ \sin y-\sin z=z-y \\ x-y+z=\pi\end{array}\right.$.
Let's write the system as $\left\{\begin{array}{c}\sin y+y=\sin x+x \\ \sin y+y=\sin z+z \\ x-y+z=\pi\end{array}\right.$. Notice that the function $f(t)=\sin t+t$ is strictly increasing.

Indeed, $f^{\prime}(t)=\cos t+1 \geq 0$; moreover, the derivative equals 0 at $t=\pi+2 k \pi, k \in \mathbf{Z}$, meaning there are no intervals where $f^{\prime}(t)=0$. Therefore, each value of the function $f(t)$ is taken at only one value of the variable. This means $x=y=z, x-y+z=\pi$, from which we get $x=y=z=\pi$.

## Answer

$(\pi, \pi, \pi)$