Which angles satisfy the following system of equations between 0 and $2 \pi$:

$$
\begin{aligned}
& \sin x+\cos y=0 \ldots \\
& \cos x \sin y=-\frac{1}{2} \ldots
\end{aligned}
$$
According to equation 1)

$$
\sin x+\cos y-\sin x+\sin \left(\frac{\pi}{2}-y\right)=\sin x-\sin \left(y-\frac{\pi}{2}\right)=0
$$

thus

$$
\sin x=\sin \left(y-\frac{\pi}{2}\right)
$$

At this point, two cases are possible.
I. $x=y-\frac{\pi}{2} ;$ thus $\cos x=\cos \left(y-\frac{\pi}{2}\right)=\cos \left(\frac{\pi}{2}-y\right)=\sin y$.

If we substitute $\sin y$ for $\cos x$ in 2), we get $\sin ^{2} y=-\frac{1}{2}$. This has no solution.

II. $x=\pi-\left(y-\frac{\pi}{2}\right)=\frac{3 \pi}{2}-y$. Now $\cos x=-\cos \left(y-\frac{\pi}{2}\right)=\sin y$. Substituting this into 2), we get: $-\sin ^{2} y=-\frac{1}{2}$

thus

$$
\sin ^{2} y=\frac{1}{2} \text { and } \sin y=+\sqrt{\frac{1}{2}}= \pm \frac{\sqrt{2}}{2} \text {. }
$$

Accordingly, if $\sin y=+\frac{\sqrt{2}}{2}$, then $y=\frac{\pi}{4}$ and $\frac{3 \pi}{4}$. When $y=\frac{\pi}{4}$, $x=\frac{3 \pi}{2}-\frac{\pi}{4}=\frac{5 \pi}{4} ;$ if $y=\frac{3 \pi}{4}$, then $x=\frac{3 \pi}{2}-\frac{3 \pi}{4}=\frac{3 \pi}{4}$.

If, however, $\sin y=-\frac{\sqrt{2}}{2}$, then $y=\frac{5 \pi}{4}$, or $\frac{7 \pi}{4}$.

In the former case, $x=\frac{3 \pi}{2}-\frac{5 \pi}{4}=\frac{\pi}{4}$, in the latter case $x=\frac{3 \pi}{2}-\frac{7 \pi}{4}=-\frac{\pi}{4}$. However, the functions of $x=-\frac{\pi}{4}$ are the same as those of $-\frac{\pi}{4}+2 \pi=\frac{7 \pi}{4}$.

Thus, our equation is satisfied by 4 value pairs, between 0 and $2 \pi$:

$$
\begin{array}{llll}
x=\frac{\pi}{4}, & \frac{3 \pi}{4}, & \frac{5 \pi}{4}, & \frac{7 \pi}{4} \\
y=\frac{5 \pi}{4}, & \frac{3 \pi}{4}, & \frac{\pi}{4}, & \frac{7 \pi}{4}
\end{array}
$$

NB. The corresponding values are listed below each other.

$\frac{\pi}{4}$ is the absolute measure of a $45^{\circ}$ angle.

Imre Dudás (Fazekas Mihály r. VI. o. Debrecen).

Note. Our equations show that $\sin x$ and $\cos y$, as well as $\cos x$ and $\sin y$, have opposite signs. This opposition of signs is found in the second and fourth quadrants, so if $x$ is in the second or fourth quadrant $\left(\frac{3 \pi}{4}\right.$, or $\left.\frac{7 \pi}{4}\right)$, then $y$ is also in the second or fourth quadrant.

In the first and third quadrants, the signs of sine and cosine are the same, both positive in the first quadrant, and both negative in the third quadrant. Therefore, if $x$ is in the first quadrant, then $y$ is in the third quadrant, and vice versa.