It is necessary to find the number of $x+y+z$ (sum of roots of equation) in the range $\left(\frac{-5\pi}{4}; \frac{13\pi}{2} \right)$:
$$ \left\{ \begin{array}{c} \sin x+\sin y=2\cos z\\ \sin y + \sin z= 2\cos x \\ \sin z + \sin x = 2 \cos y \end{array} \right. $$ I tried different ways, trying to simplify, but even couldn't find real roots.
If you are interested in only one solution, choose $x=y=z$, simplifying to only one equation:
$\sin(x)=\cos(x)$
which is e.g. satisfied for $x=\frac{\pi}{4}$. Your soultion would then be $x+y+z=3x=\frac{3\pi}{4}$.