Solving a system of equations with trig

You are on the right track. Subtracting $(2)$ from $(1)$ gives $\sin x=c+\sin y$ where $c=\frac{a-b}3$ and putting this into $(1)$ gives $$a=2c+\sin y+\sin x\cos y+\sin y\cos x$$ so $$\cos x=-1-\cos y+\frac{b+c-c\cos y}{\sin y}$$ Thus $$\small \sin^2x+\cos^2x\\=\\\small c^2+2c\sin y+\sin^2y+1+\cos^2y+\left(\frac{b+c-c\cos y}{\sin y}\right)^2+2\cos y-2\frac{b+c-c\cos y}{\sin y}-2\cos y\frac{b+c-c\cos y}{\sin y}$$ so $$\left(\frac{b+c-c\cos y}{\sin y}-1\right)^2-2\cos y\left(\frac{b+c-c\cos y}{\sin y}-1\right)+c^2+2c\sin y=0$$ giving $$\frac{b+c-c\cos y}{\sin y}-1=\frac{2\cos y\pm\sqrt{4\cos^2y-4(c^2+2c\sin y)}}2$$ and the half-angle identities give $$\frac b{\sin y}+c\tan\frac y2=2\cos^2\frac y2\pm\sqrt{1-(c+\sin y)^2}$$ I highly doubt there is an analytical solution to this, but once $y$ is found perhaps numerically, $x=\sin^{-1}(c+\sin y)$.