Solve the following equation:

$$
\cos 3 x+2 \cos x=0
$$
According to a known formula

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

substituting this value into the equation

$$
4 \cos ^{3} x-3 \cos x+2 \cos x=0
$$

or

$$
\cos x \cdot\left(4 \cos ^{2} x-1\right)=0
$$

It follows that either

$$
\cos x=0, \quad \text { from which } \quad x_{1,2}=\frac{\pi}{2} \pm k \pi \quad(k=0,1,2, \ldots)
$$

or

$$
4 \cos ^{2} x-1=0
$$

from which

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

and thus

$$
x_{3,4}=\frac{\pi}{3} \pm k \pi \quad(k=0,1,2, \ldots) \quad \text { or } \quad x_{5,6}=\frac{2 \pi}{3} \pm k \pi \quad(k=0,1,2, \ldots)
$$

Therefore, there are a total of 6 roots in the interval $(0,2 \pi)$.

Pál Bártfai (Bp., I., Petőfi g. III. o. t.)

Many solutions were not acceptable because the solver divided by $\cos x$ without further ado and thus lost the $x_{1,2}$ roots.