Trigonometric system 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) $

Multiply the 2nd equation by $i$ and add the 3rd equation to get $$e^{ix}+e^{iy}+e^{iz}=e^{i\pi/3}$$ Now you can argue geometrically that for three numbers of modulus 1 to sum to a number of modulus 1, one of the summands must be the negative of another summand (and so the third summand must equal the lone number on the other side). From there, it's easy. 
If you don't like geometry, you can rewrite as $$e^{ix}+e^{iy}=e^{i\pi/3}-e^{iz}$$ then express the left side as $e^{is}\cos t$ and the right side as $e^{iu}\sin v$ for appropriate $s,t,u,v$, and that should give you enough relations among $x,y,z$ to solve the problem. 
If you haven't learned about complex exponentials, the above won't mean much to you - but the upside is that you have some very nice mathematics to look forward to. 
