$\sin^{-1}(\sin x)$ is a multi-valued function reduces to $n\pi+(-1)^nx$ where $n$ is any integer.
So, $n\pi+(-1)^nx+m\pi+(-1)^mx=x+y$
(1)If $m,n$ both are even, $(m+n)\pi=0\implies m+n=0$
We can not find any relation between $x,y$ here.
(2)If $m$ is odd$=2M+1$(say), and $n$ is even$=2N$, $2N\pi+x+(2M+1)\pi-y=x+y\implies 2y=(2N+2M+1)\pi\implies y=\frac{(2N+2M+1)\pi}{2}=(N+M)\pi+\frac{\pi}{2}\implies y=-3\frac{\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$
(3)Similarly if $m$ is even, $n$ is odd, $x=-3\frac{\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$
(4) If both are odd, let $m=2M+1,n=2N+1, (2N+1)\pi-x+(2M+1)\pi-y=x+y$ $\implies x+y=(M+N+1)\pi$ Now, $-2\pi ≤x+y ≤2\pi$
So, $x+y=±2\pi, ±\pi, 0$.
Alternatively, the principal value of $sin^{-1}(\sin x)$ must lie $\in [-\frac{\pi}{2};\frac{\pi}{2}]$
Let $sin^{-1}(\sin x)=z\implies \sin z=\sin x$
$z=x$ if $-\frac{\pi}{2}≤x≤\frac{\pi}{2}$(as Thomas Andrews has observed),
$z=\pi-x$ if $\frac{\pi}{2}
$z=-x-\pi$ if $-\pi≤x<-\frac{\pi}{2}$
So, there are three regions for $x$.
So, there are $3\cdot 3=9$ regions for $x,y$.
For example,
If $-\frac{\pi}{2}≤x≤\frac{\pi}{2}$ and $-\frac{\pi}{2}≤y≤\frac{\pi}{2}$, any values of $x,y$ in the above ranges will satisfy the given equation.
If $-\frac{\pi}{2}≤x≤\frac{\pi}{2}$ and $\frac{\pi}{2} $x+\pi-y=x+y\implies y=\frac{\pi}{2}$ which is impossible as $\frac{\pi}{2}
If $-\frac{\pi}{2}≤x≤\frac{\pi}{2}$ and $-\pi≤y<-\frac{\pi}{2},$ $x-y-\pi=x+y\implies y=-\frac{\pi}{2}$ which is again impossible
and so on.