
1. Find all real solutions of the system

$$
\begin{aligned}
& 2 \sin x \cos (x+y)+\sin y=1 \\
& 2 \sin y \cos (y+x)+\sin x=1
\end{aligned}
$$


Solution. Using the familiar formulas

$$
\cos (x+y)=\cos x \cos y-\sin x \sin y, \quad \sin 2 x=2 \sin x \cos x, \quad \cos 2 x=1-2 \sin ^{2} x
$$

we can rewrite the left-hand side of the first equation as

$$
\begin{aligned}
2 \sin x \cos (x+y)+\sin y & =2 \sin x(\cos x \cos y-\sin x \sin y)+\sin y= \\
& =2 \sin x \cos x \cos y+\left(1-2 \sin ^{2} x\right) \sin y= \\
& =\sin 2 x \cos y+\cos 2 x \sin y= \\
& =\sin (2 x+y) .
\end{aligned}
$$

Similarly the left-hand side of the second equation equals $\sin (2 y+x)$. The given system is thus equivalent to

$$
\begin{aligned}
& \sin (2 x+y)=1 \\
& \sin (2 y+x)=1
\end{aligned}
$$

Since the sine function assumes the value 1 precisely at points of the form $\frac{1}{2} \pi+$ $2 k \pi$, where $k$ is an integer, the solutions of the last system will be precisely those pairs $(x, y)$ for which there exist integers $k, l$ such that

$$
2 x+y=\frac{1}{2} \pi+2 k \pi, \quad 2 y+x=\frac{1}{2} \pi+2 l \pi .
$$

Multiplying the first equation by two and subtracting the second (or expressing one of the variables from the first equation and substituting it into the second), we obtain after a small manipulation

$$
x=\frac{1}{6} \pi+\frac{2}{3}(2 k-l) \pi, \quad y=\frac{1}{6} \pi+\frac{2}{3}(2 l-k) \pi .
$$

The solutions of the system are therefore the pairs $\left(\frac{1}{6} \pi+\frac{2}{3}(2 k-l) \pi, \frac{1}{6} \pi+\frac{2}{3}(2 l-k)\right)$, where $k, l$ are arbitrary integers. Verification is not necessary, since it is clear from the argument that these pairs satisfy the relations (2), and, hence, also the system (1), which is equivalent to the original system.

Remark. The result can be written also in a different way: since $y-x=\frac{1}{3}(6 l-$ $6 k) \pi=2(l-k) \pi$, setting $m=l-k, n=2 k-l$ we can write $x=\frac{1}{6} \pi+\frac{2}{3} n \pi$, $y=x+2 m \pi$, so the solutions are the pairs $\left(\frac{1}{6} \pi+\frac{2}{3} n \pi, \frac{1}{6} \pi+\frac{2}{3} n \pi+2 m \pi\right)$, where $m$, $n$ are arbitrary integers. (If $k, l$ range over all possible pairs of integers, then so do $m, n$.)
