I would like to solve:
$ x +y+z=\frac{11\pi}{6} $
$ \sin(x)+\sin(y)+\sin(z)= \frac{\sqrt{3}}{2} $
$ \cos(x)+\cos(y)+\cos(z)=\frac{1}{2} $
After eliminating $ z $ I get:
$ 2\sin(x)+2\sin(y)-\cos(x+y)-\sqrt{3}\sin(x+y)=\sqrt{3}\tag{1}$
$ 2\cos(x)+2\cos(y)+\sqrt{3}\cos(x+y)-\sin(x+y)=1\tag{2}$
Also: $\sqrt{3}\times(1)+(2)$ gives $ \cos(x-\frac{\pi}{3})+\cos(y-\frac{\pi}{3})-\sin(x+y)=1 $...
My attempt:
$ \sin(x)+\sin(y)+\sin(z)=\sin(x+y+z+\frac{\pi}{2}) $
$ \sin(x)+\sin(y)+\sin(z)-\sin(x+y+z+\frac{\pi}{2})=0 $
$(1)$: $ \sin(\frac{x+y}{2})\cos(\frac{x-y}{2})-\sin(\frac{x+y}{2}+\frac{\pi}{4})\cos(\frac{x+y+2z}{2}+\frac{\pi}{4})=0 $
$ \cos(x)+\cos(y)+\cos(z)-\cos(x+y+z+\frac{\pi}{2})=0 $
$ (2) $: $ \cos(\frac{x+y}{2})\cos(\frac{x-y}{2})+\sin(\frac{x+y+2z}{2}+\frac{\pi}{4})\sin(\frac{x+y}{2}+\frac{\pi}{4})=0 $
$(1)+(2)$: $ \cos(\frac{x-y}{2})\cos(\frac{x+y}{2}-\frac{\pi}{4})+\sin(\frac{x+y}{2}+\frac{\pi}{4})\sin(\frac{x+y+2z}{2})=0 $
$ \cos(\frac{x+y}{2}-\frac{\pi}{4})(\cos(\frac{x-y}{2})+\sin(\frac{x+y+2z}{2}))=0 $
$ \cos(\frac{x+y}{2}-\frac{\pi}{4})(\sin(\frac{x-y}{2}+\frac{\pi}{2})+\sin(\frac{x+y+2z}{2}))=0 $
$ \cos(\frac{x+y}{2}-\frac{\pi}{4})\sin(\frac{x+z}{2}+\frac{\pi}{4})\cos(\frac{y+z}{2}-\frac{\pi}{4})=0 $
$ x+y=-\frac{\pi}{2} (\mod2\pi) $
$ x+z=-\frac{\pi}{2} (\mod2\pi) $
$ y+z=-\frac{\pi}{2} (\mod2\pi) $
Using only these equations to determine $ x,y,z $:
$ x=y=z=-\frac{\pi}{4} (\mod\pi) $ and the system is not satisfied.
Using $ x+y+z=\frac{11\pi}{6} $ to determine one of the three quantities:
$x=y=-\frac{\pi}{4} (\mod\pi) $, $ z=\frac{\pi}{3} (\mod\pi) $ or $x=z=-\frac{\pi}{4} (\mod\pi) $, $ y=\frac{\pi}{3} (\mod\pi) $ or $y=z=-\frac{\pi}{4} (\mod\pi) $, $ x=\frac{\pi}{3} (\mod\pi) $