Solve the following system of equations:

$$
\begin{gathered}
\frac{\sin x+\cos x}{\sin y+\cos y}+\frac{\sin y-\cos y}{\sin x+\cos x}=\frac{1}{\sin (x+y)+\cos (x-y)} \\
2(\sin x+\cos x)^{2}-\left(2 \cos ^{2} y+1\right)=\frac{\sqrt{3}}{2}
\end{gathered}
$$
The first equation in the task is meaningless if any of the denominators is 0, that is, if

$$
\sin x+\cos x=0
$$

which means

$$
\begin{aligned}
& x_{1}=135^{\circ} \pm n \cdot 360^{\circ} \\
& x_{2}=315^{\circ} \pm n \cdot 360^{\circ}
\end{aligned}
$$

Similarly, from the second denominator, it is 0 if

$$
y_{1}=135^{\circ} \pm n \cdot 360^{\circ}, \quad y_{2}=315^{\circ} \pm n \cdot 360^{\circ}
$$

Based on the relationships involving the sum of two angles, the third denominator can be factored into the product of the first two denominators:

$$
\sin (x+y)+\cos (x-y)=(\sin x+\cos x)(\sin y+\cos y)
$$

thus, it is also 0 when the first two are.

Excluding the mentioned roots, the equation can be multiplied throughout by the least common multiple of the denominators:

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

Substituting $1-\cos ^{2} y$ for $\sin ^{2} y$:

$$
(\sin x+\cos x)^{2}-2 \cos ^{2} y=0
$$

Subtract this equation from the second equation of the system:

$$
(\sin x+\cos x)^{2}=\frac{\sqrt{3}}{2}+1
$$

Expanding the left side. Using the relationships $\sin ^{2} x+\cos ^{2} x=1$ and $2 \sin x \cos x=\sin 2 x$, we get the following equation:

$$
\sin 2 x=\frac{\sqrt{3}}{2}
$$

From this,

$$
\begin{array}{lll}
2 x_{1}=60^{\circ} \pm n \cdot 360^{\circ}, & \text { i.e. } & x_{1}=30^{\circ} \pm n \cdot 180^{\circ} \\
2 x_{2}=120^{\circ} \pm n \cdot 360^{\circ}, & \text { i.e. } & x_{2}=60^{\circ} \pm n \cdot 180^{\circ}
\end{array}
$$

$(n=0,1,2,3, \ldots)$

The simplest way to determine $y$ is to subtract twice (1) from the second equation of the system. This leads to the following equation:

$$
\begin{aligned}
& 2 \cos ^{2} y=\frac{\sqrt{3}}{2}+1 \\
& \cos y= \pm \sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}
\end{aligned}
$$

Since $\cos \alpha= \pm \sqrt{\frac{1+\cos 2 \alpha}{2}}$ and ignoring multiples of $360^{\circ}$,

$$
\cos 30^{\circ}=\cos 330^{\circ}=\frac{\sqrt{3}}{2}
$$

therefore

$$
2 y_{1}=30^{\circ} \pm n \cdot 360^{\circ}
$$

and

$$
2 y_{2}=330^{\circ} \pm n \cdot 360^{\circ}
$$

which means

$$
y_{1}=15^{\circ} \pm n \cdot 180^{\circ}
$$

$$
y_{2}=165^{\circ} \pm n \cdot 180^{\circ}
$$

None of the excluded roots were obtained as solutions. By substitution, we can verify that the obtained roots are indeed solutions to the system of equations.

Bollobás Béla (Bp. V., Apáczai Csere g. I. o. t.)