By adding the equations and simplifying we get $\sin x = -\sin y$. Thus $y = -x + 2k\pi$ or $y = \pi + x + 2k\pi = x + (2k + 1)\pi$. In the second case we get that $|y - x| = |(2k + 1)\pi| \ge \pi$, but from the first equality $|y - x| = |\sin x| \le 1 < \pi$, a contradiction. Therefore $y = -x + 2k\pi$, where $k$ is an integer. By plugging this into the first equation we get $2x + \sin x = 2k\pi$ after simplifying. We see that this is satisfied in case of any integer $k$ by the value $x = k\pi$; then also $y = -k\pi + 2k\pi = k\pi$. As $f(x) = 2x + \sin x$ is an increasing function, there cannot be any other solutions to $2x + \sin x = 2k\pi$.

Function $f(z) = z + \sin z$ is strictly increasing, because its derivative $f'(z) = 1 + \cos z$ is positive everywhere, except for some isolated points. Therefore if $x < y$ in case of some solution $(x, y)$ to the system of equations, then $y = x + \sin x < y + \sin y = x$, a contradiction. Analogously $y < x$ gives a contradiction. In conclusion $x = y$ is the only option. By substituting it in we get $\sin x = \sin y = 0$, from where $x = y = k\pi$ for any integer $k$. All pairs $(k\pi, k\pi)$ indeed satisfy the given system of equations.