Solve the following system of equations:

$\cos x+\cos y=\cos (x+y)$,

$\sin x+\sin y=\sin (x+y)$.
Reducing our equations to 0, based on known identities,

$$
\begin{gathered}
2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}-\left(2 \cos ^{2} \frac{x+y}{2}-1\right)= \\
=2 \cos \frac{x+y}{2}\left(\cos \frac{x-y}{2}-\cos \frac{x+y}{2}\right)+1=0 \\
2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}-2 \sin \frac{x+y}{2} \cos \frac{x+y}{2}= \\
=2 \sin \frac{x+y}{2}\left(\cos \frac{x-y}{2}-\cos \frac{x+y}{2}\right)=0
\end{gathered}
$$

If the second factor were 0 here, (1') could not be satisfied, so we only need to consider the case where the first factor turns out to be zero:

$$
\sin \frac{x+y}{2}=0, \quad \frac{x+y}{2}=k \cdot \pi, \quad x+y=k \cdot 2 \pi, \quad y=k \cdot 2 \pi-x
$$

If (3) holds, then 0 stands on both sides of (2), and (1) becomes

$$
2 \cos x=1
$$

indeed, $\cos y=\cos (-x)=\cos x$. From this,

$$
\cos x=\frac{1}{2}, \quad x= \pm \frac{\pi}{3}+2 m \pi \quad(m \text { integer })
$$

and based on (3),

$$
y=\mp \frac{\pi}{3}+2 n \pi \quad(n \text { integer })
$$

We performed equivalent transformations everywhere, so there is no need to verify the roots.

Poór Zsolt (Bátaszék, Gymn., IV. grade)