Solve the following equation:

$$
2 \cos x+\cos 3 x+\cos 5 x=0
$$
The given equation can also be written as:

$$
2 \cos x+\cos (4 x-x)+\cos (4 x+x)=0
$$

or

$$
\begin{gathered}
2 \cos x+\cos 4 x \cos x+\sin 4 x \sin x+ \\
+\cos 4 x \cos x-\sin 4 x \sin x=0 \\
2 \cos x+2 \cos 4 x \cos x=0 \\
\cos x(1+\cos 4 x)=0 \\
2 \cos x \cos ^{2} 2 x=0
\end{gathered}
$$

from which

$$
\cos x=0 \text { and } \cos 2 x=0
$$

and thus

$$
x=k \cdot 90^{\circ} \text { and } x=k \cdot 45^{\circ}
$$

where $k$ is any odd number, or if $n$ is any integer not divisible by 4, then $x=n \cdot 45^{\circ}$.

(Júlia Póka, Losoncz.)

Number of solutions: 44.