Solve the following system of equations ( $x$ and $y$ are in radians):

$$
\begin{gathered}
\sin x+\sin y=\sin (x+y), \\
|x|+|y|=1 .
\end{gathered}
$$
(1) Writing the expressions on the left and right side as a product, and then transforming, we get

$$
2 \sin \frac{x+y}{2}\left(\cos \frac{x-y}{2}-\cos \frac{x+y}{2}\right)=0
$$

Since the expression in the parentheses is equal to $\left(2 \sin \frac{x}{2} \cdot \sin \frac{y}{2}\right)$, (1) holds if and only if

$$
\text { either } \quad \frac{x+y}{2}=k_{1} \pi, \quad \text { or } \quad \frac{x}{2}=k_{2} \pi, \quad \text { or } \quad \frac{y}{2}=k_{3} \pi
$$

$\left(k_{1}, k_{2}, k_{3}\right)$ are integers. However, according to (2),

$$
|x|<\pi, \quad|y|<\pi \quad \text { and } \quad|x+y| \leqq|x|+|y|<\pi
$$

which means only $k_{1}=k_{2}=k_{3}=0$ is possible. Thus, depending on which of the above three cases follows, the system of equations has the following six solutions:

| $\mathrm{x}$ | $-0.5$ | 0.5 | 0 | 0 | 1 | -1 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{y}$ | 0.5 | $-0.5$ | 1 | -1 | 0 | 0 |

Frits Gabriella (Sopron, Széchenyi I. Gymnasium)