The left side can be written using the identity $\sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}$ as follows:

$$
(\sin 3 x+\sin x)+\sin 2 x=2 \sin 2 x \cos x+\sin 2 x=\sin 2 x(2 \cos x+1)
$$

The right side can be transformed using the identities $\sin ^{2} \alpha+\cos ^{2} \alpha=1$ and $\cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alpha$ as follows: $(1+\cos 2 x)+\cos x=2 \cos ^{2} x+\cos x=\cos x(2 \cos x+1)$. Reducing our equation to 0, we get

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

or

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

that is, substituting $2 \sin x \cos x$ for $\sin 2 x$

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

Thus, either

$$
\begin{aligned}
& \cos x=0, \text { or } \\
& 2 \cos x+1=0, \text { or } \\
& 2 \sin x-1=0
\end{aligned}
$$

From (1)

$$
x_{1,2}=90^{\circ} \pm k \cdot 180^{\circ}
$$

From (2)

$$
\cos x=-\frac{1}{2}
$$

from which

$$
x_{3}=120^{\circ} \pm k \cdot 360^{\circ}, \quad x_{4}=240^{\circ} \pm k \cdot 360^{\circ}
$$

From (3)

$$
\sin x=\frac{1}{2}
$$

from which

$$
x_{5}=30^{\circ} \pm k \cdot 360^{\circ}, \quad x_{6}=150^{\circ} \pm k \cdot 360^{\circ}
$$

where $k=0,1,2, \ldots$

By substitution, we can verify that all six roots satisfy our equation.

Ádám Antal (Bp., VIII., Széchenyi g. III. o. t.)
Solve the following equation:

$$
\sin x + \sin 2x + \sin 3x = 1 + \cos x + \cos 2x
$$