Solve the following equation:

$$
2 \sin 3 x=3 \cos x+\cos 3 x
$$
First solution.

$$
\begin{gathered}
\sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\cos 2 x \sin x= \\
=2 \sin x \cos ^{2} x+\left(2 \cos ^{2} x-1\right) \sin x=4 \sin x \cos ^{2} x-\sin x
\end{gathered}
$$

Similarly,

$$
\begin{gathered}
\cos 3 x=\cos 2 x \cos x-\sin 2 x \sin x= \\
=\left(1-2 \sin ^{2} x\right) \cos x-2 \sin ^{2} x \cos x=\cos x-4 \sin ^{2} x \cos x
\end{gathered}
$$

Thus, our equation becomes:

$$
8 \sin x \cos ^{2} x+4 \sin ^{2} x \cos x-4 \cos x-2 \sin x=0
$$

which simplifies to

$$
=2 \cos x(2 \sin x \cos x -1)+\sin x(2 \sin x \cos x-1)=0
$$

or

$$
(\sin 2 x-1)(2 \cos x+\sin x)=0
$$

If

$$
\sin 2 x-1=0
$$

then

$$
x_{1}=45^{\circ} \pm k \pi
$$

If

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

then we can divide by $\cos x$, since $x=90^{\circ}$ is not a root, so

$$
\operatorname{tg} x=-2
$$

thus

$$
x_{2}=116^{\circ} 33^{\prime} 54^{\prime \prime} \pm k \pi
$$

(Rosenberg Jenö, Keszthely.)

Second solution.

$$
\sin 3 x=\sin (2 x+x)=3 \sin x-4 \sin ^{3} x
$$

and

$$
\cos 3 x=4 \cos ^{3} x-3 \cos x
$$

so our equation becomes

$$
6 \sin x-8 \sin ^{3} x=4 \cos ^{3} x
$$

$x=90^{\circ}$ is not a root of our equation, so we can divide every term by $2 \cos ^{3} x$:

$$
3 \operatorname{tg} x \cdot \frac{1}{\cos ^{2} x}-4 \operatorname{tg}^{3} x=2
$$

But

$$
\frac{1}{\cos ^{2} x}=\sec ^{2} x=1+\operatorname{tg}^{2} x
$$

so

$$
\operatorname{tg}^{3} x-3 \operatorname{tg} x+2=0
$$

This can be written as:

$$
(\operatorname{tg} x-1)\left(\operatorname{tg}^{2} x+\operatorname{tg} x-2\right)=0
$$

If

$$
\operatorname{tg} x-1=0
$$

then

$$
x_{1}=45^{\circ} \pm k \pi
$$

If

$$
\operatorname{tg}^{2} x+\operatorname{tg} x-2=0
$$

then

$$
\operatorname{tg} x=\frac{-1 \pm 3}{2}
$$

thus

$$
x_{2}=116^{\circ} 33^{\prime} 54^{\prime \prime} \pm k \pi
$$

(Haar Alfréd, Budapest.)

The problem was also solved by: Dömény I., Fekete M., Jánosy Gy., Krampera Gy., Kürti I., Messer P., Pám M., Riesz M., Schöffer I., Sonnenfeld J., the V. district math. circle.