\left\{\begin{matrix}
x+y=\pi\\
\sin (\pi -x)+ \sqrt3 \cos y =1
\end{matrix}\right. \to
\left\{\begin{matrix}
y=\pi-x\\
\sin (\pi -x)+ \sqrt3 \cos (\pi-x) =1
\end{matrix}\right.\\\\
da qui poi:
\sin (\pi -x)+ \sqrt3 \cos (\pi-x) =1 \to \sin x -\sqrt3 \cos x=1 \to \\
\sqrt3 \cos x=\sin x -1 \to \sqrt3 \sqrt{1-\sin^2x}=\sin x -1\to\\
3(1-\sin^2x)=\sin^2x+1-2\sin x \to 3-3\sin^2x=\sin^2x+1-2\sin x \to\\
4\sin^2x-2\sin x-2=0 \to 2\sin^2x-\sin x-1=0 \to\\
\sin x= \frac{1\pm\sqrt{1+8}}{4}=\frac{1\pm3}{4} \to \sin x=-\frac{1}{2}, \sin x=1\\\
y=\pi-x=...
Da qui puoi ricavarti facilmente le x e poi ricavare le y di conseguenza.
