$\sin2x-a\sin\left(x-b\right)=\sin\left(x+y\right)$
$\sin2y-a\sin\left(y-b\right)=\sin\left(x+y\right)$
$a$ and $b$ are constant values,
$x$ and $y$ are unknown variables .
first substract them:
$2\cos\left(\frac{x-y}{2}\right)\cos\left(x+y\right)=a\cos\left(\frac{x+y}{2}-b\right)$
and adding them:
$2\left(1-\cos^{2}\left(\frac{x-y}{2}\right)\right)\sin\left(x+y\right)+a\sin\left(\frac{x+y}{2}-b\right)\cos\left(\frac{x-y}{2}\right)=0$
now multiply this equation above with $\cos^{2}\left(x+y\right)$, we get :
$2\left(\cos^{2}\left(x+y\right)-\cos^{2}\left(\frac{x-y}{2}\right)\cos^{2}\left(x+y\right)\right)\sin\left(x+y\right)+a\sin\left(\frac{x+y}{2}-b\right)\cos\left(\frac{x-y}{2}\right)\cos^{2}\left(x+y\right)=0$
now replace $2\cos\left(\frac{x-y}{2}\right)\cos\left(x+y\right)$ with $a\cos\left(\frac{x+y}{2}-b\right)$,we get :
$\left(2\cos^{2}\left(x+y\right)-\frac{1}{2}a^{2}\cos^{2}\left(\frac{x+y}{2}-b\right)\right)\sin\left(x+y\right)+\frac{1}{4}a^{2}\sin\left(x+y-2b\right)\cos\left(x+y\right)=0$
now we have only one variable in this equation ,which is $\frac{x+y}{2}$ .
then we can simplify it down to :
$8z^{3}-8z+a^{2}z+a^{2}\sin2b=0$
here $z=\sin\left(x+y\right)$
now solving this equation is trival, thus problem solved .