2. Solve the system of equations:

$$
\left\{\begin{array}{l}
\sin x \cos y=-\frac{1}{2} \\
\cos x \sin y=\frac{1}{2}
\end{array}\right.
$$
Solution. By collecting the two equations and subtracting the first equation from the second, we get

$$
\left\{\begin{array}{l}
\sin (x+y)=0 \\
\sin (y-x)=1
\end{array}\right.
$$

Then: $x+y=n \pi, y-x=\frac{\pi}{2}+2 k \pi$, from which it follows that:

$$
x=\pi\left(\frac{n}{2}-k-\frac{1}{4}\right), \quad y=\pi\left(\frac{n}{2}+k+\frac{1}{4}\right), n, k \in \mathbb{Z}
$$