Let's determine the values of $\sin x$ and $\sin y$ if

$$
\begin{gathered}
\sin y=k \sin x \ldots \\
2 \cos x+\cos y=1 \ldots
\end{gathered}
$$

For which values of $k$ does a solution exist?
From (2) $\cos y=1-2 \cos x$.

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

$$
4 \cos ^{2} x-4 \cos x+k^{2}\left(1-\cos ^{2} x\right)=0 \quad \text { or } \quad\left(4-k^{2}\right) \cos ^{2} x-4 \cos x+k^{2}=0 \ldots
$$

From this, $\cos x=\frac{4 \pm \sqrt{16-4 k^{2}\left(4-k^{2}\right)}}{2(4-k)^{2}}=\frac{4 \pm \sqrt{4 k^{4}-16 k^{2}+16}}{2\left(4-k^{2}\right)}=\frac{4 \pm\left(2 k^{2}-4\right)}{2\left(4-k^{2}\right)}$.

$$
\cos x_{1}=\frac{k^{2}}{4-k^{2}}, \quad \cos x_{2}=\frac{2\left(4-k^{2}\right)}{2\left(4-k^{2}\right)}=1
$$

The value of $\cos x_{1}$ is acceptable if its absolute value is not greater than 1, that is:

$$
k^{4} \leq\left(4-k^{2}\right)^{2} \quad \text { or } \quad k^{2} \leq 2, \quad \text { thus } \quad-\sqrt{2} \leq k \leq+\sqrt{2}
$$

In this case, $\cos y_{1}=1-\frac{2 k^{2}}{4-4 k^{2}}=\frac{4-3 k^{2}}{4-k^{2}}$

This value of $\cos y_{1}$ is valid because, if $k^{2} \leq 2$, it can be easily shown that

$$
-1 \leq \frac{4-3 k^{2}}{4-k^{2}}<1
$$

When $\cos x_{2}=1$, $\sin x_{2}=0$, so $\sin y_{2}=0, \cos y_{2}=-1$, thus

$$
x_{2}=2 k \pi, \quad y_{2}=(2 l+1) \pi \text {. }
$$

László Ballay (Bencés g. VI. o., Győr).