1. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& 2 \sin x \cos (x+y)+\sin y=1 \\
& 2 \sin y \cos (y+x)+\sin x=1
\end{aligned}
$$
SOLUTION. Using the known 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 obtain by transforming the left side of the first equation

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

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

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

Since the sine function takes 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 system will be exactly those pairs $(x, y)$ for which there exist integers $k, l$ such that

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

From this, by subtracting appropriate multiples of the equations (e.g., by subtracting the second from twice the first), or by directly expressing one variable from the first equation and substituting into the second equation after rearrangement, we get

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

The solutions of the system are thus 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 any integers. There is no need to verify, as the given procedure shows that such pairs $(x, y)$ satisfy the relations $(2)$, and thus the system (1), which is equivalent to the given system.

Note. The result can be written differently. Since $y-x=$ $=\frac{1}{3}(6 l-6 k) \pi=2(l-k) \pi$, we can write $x=\frac{1}{6} \pi+\frac{2}{3} n \pi$, $y=x+2 m \pi$, where $m= l-k, n=2 k-l$. The solutions are thus 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 any integers. (If $k, l$ run through all possible pairs of integers, the same is true for the numbers $m, n$.)

ANOTHER SOLUTION. If a pair $(x, y)$ is a solution of the given system, then due to the periodicity of the sine and cosine functions with period $2 \pi$, all pairs $(x+2 k \pi, y+2 l \pi)$ $(k, l$ are any integers) are also solutions. We will therefore solve the system only in the interval $\langle 0,2 \pi)$ and at the end, shift the found solutions by $(2 k \pi, 2 l \pi)$ to obtain the general solution.

By subtracting the equations of the system, we obtain after factoring the left side the equation

$$
(\sin x-\sin y)(2 \cos (x+y)-1)=0 .
$$

We distinguish two cases depending on which of the factors is zero.

I. If $\sin x=\sin y$, we have, given the condition $x, y \in\langle 0,2 \pi)$, three possibilities: either $x=y$, or $x+y=\pi$, or $x+y=3 \pi$ (see Fig. 1).

![](https://cdn.mathpix.com/cropped/2024_04_17_9bf59b031063fd37bc34g-02.jpg?height=371&width=1262&top_left_y=797&top_left_x=403)

Fig. 1

For $x=y$, substituting into the original system, we get the single equation

$$
2 \sin x \cos 2 x+\sin x=1
$$

From this, using the formula $\cos 2 x=1-2 \sin ^{2} x$ and substituting $\sin x=t$, we get by equivalent transformations

$$