[Trigonometric Systems of Equations and Inequalities]

Find all solutions to the system of equations

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

satisfying the conditions $0 \leq x \leq \pi, ; ; 0 \leq y \leq \pi$.
From the first equation, we get $x+y=k \pi$. From the second equation, $x-y=k_{1} \pi$. Therefore, $x=\left(\left(k+k_{1}\right) \pi\right) / 2, y=\left(\left(k-k_{1}\right) \pi\right) / 2$. Since we need to choose values for $x$ and $y$ within the range from 0 to $\pi$, then

$$
\left\{\begin{array}{l}
0 \leq k+k_{1} \leq 2 \\
0 \leq k-k_{1} \leq 2
\end{array}\right.
$$

From this system of inequalities, we find the possible values of $k$ and $k_{1} \cdot k$ can have values $0,1,2$. When $k=0, k_{1}$ can have the value 0; when $k=1, k_{1}=-1,0,1$; when $k=2, k_{1}=0$.

## Answer

$(0,0),(0, \pi),(\pi / 2, \pi / 2),(\pi, 0),(\pi, \pi)$.