Explicit solution of equations with trigonometric function

Question

How can an explicit solution for $x,y$ be obtained given $$a\sin(x-y)=\sin x$$ $$a\sin(y-x)=\sin y$$

Answer 1

Obviously:

$$-\sin x=\sin y$$

It means that:

$$y=x + \pi + 2k\pi \tag{1}$$

...or:

$$y=-x+2k\pi\tag{2}$$

Case (1):

$$a\sin(x-(x+\pi+2k\pi))=-\sin x$$

$$a\sin(-(2k+1)\pi)=-\sin x$$

$$0=\sin x$$

$$x=n\pi, \space y=n\pi+\pi+2k\pi=m\pi$$

...for any $m,n\in Z$

Case (2):

$$a\sin(x-(-x+2k\pi)=\sin x$$

$$a\sin(2x-2k\pi)=\sin x$$

$$a\sin(2x)=\sin x$$

$$2a\sin x \cos x=\sin x$$

Trivial solution is $\sin x=0$, but that solution is already obtained in case (1). Non-trivial solution is:

$$\cos x=\frac1{2a}$$

$$x=\pm\arccos\frac1{2a} + 2n\pi$$

$$y=-(\pm\arccos\frac1{2a} + 2n\pi) + 2k\pi$$

$$y=\mp\arccos\frac1{2a} + 2m\pi$$

...for any $m,n\in Z$. Obviously, this solution exists only for:

$$-1\le\frac1{2a}\le1$$