$$\left\{\begin{matrix} x+y=\pi\\ \sin (\pi -x)+ \sqrt3 \cos y =1 \end{matrix}\right. \to \left\{\begin{matrix} y=\pi-x\\ \sin (\pi -x)+ \sqrt3 \cos (\pi-x) =1 \end{matrix}\right.\\\\$$
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.