Solve the following equation:

$$
\sin x+\sin 2 x+\sin 3 x=4 \cos \frac{x}{2} \cos x \cos \frac{3 x}{2}
$$
The given equation can also be written as:

$$
\begin{gathered}
2 \sin \frac{x}{2} \cos \frac{x}{2} + 2 \sin \frac{5 x}{2} \cos \frac{x}{2} = 4 \cos \frac{x}{2} \cos x \cos \frac{3 x}{2} \\
2 \cos \frac{x}{2} \left( \sin \frac{x}{2} + \sin \frac{5 x}{2} \right) = 4 \cos \frac{x}{2} \cos x \cos \frac{3 x}{2} \\
4 \cos \frac{x}{2} \sin \frac{3 x}{2} \cos x - 4 \cos \frac{x}{2} \cos x \cos \frac{3 x}{2} = 0 \\
4 \cos \frac{x}{2} \cos x \left( \sin \frac{3 x}{2} - \cos \frac{3 x}{2} \right) = 0
\end{gathered}
$$

The equation is satisfied if any of the factors are zero, that is, if

$$
\cos \frac{x}{2} = 0, \cos x = 0, \sin \frac{3 x}{2} - \cos \frac{3 x}{2} = 0
$$

From the first equation:

$$
\frac{x}{2} = k \cdot 360^{\circ} \pm 90^{\circ}, \text{ or } x = (4 k \pm 1) \cdot 180^{\circ}
$$

From the second equation:

$$
x = k \cdot 360^{\circ} \pm 90^{\circ}
$$

From the third equation:

$$
\frac{3 x}{2} = k \cdot 180^{\circ} + 45^{\circ}, \text{ or } x = (4 k + 1) \cdot 30^{\circ}
$$

The problem was solved by: Bobál S., Dolowschiák M., Filkorn J., Freibauer E., Jankovich S., Kárf J., Kohn B., Kornis Ö., Krausz B., Krisztián Gy., Lukhaub Gy., Mocsonyi A., Obláth R., Perl Gy., Pollák N., Petrogalli G., Rehberger Z., Sasvári G., Spitzer Ö., Weisz J.