Solve the following equation:

$$
\frac{\operatorname{cot} \alpha}{\sin \alpha}=2 \sqrt{3}
$$
Using the relationship $\operatorname{ctg} \alpha=\frac{\cos \alpha}{\sin \alpha}$, our equation can be written as:

$$
\frac{\cos \alpha}{\sin ^{2} \alpha}=2 \sqrt{3}
$$

where we must assume that $\sin \alpha \neq 0$.

Substituting $\sin ^{2} \alpha$ with $\left(1-\cos ^{2} \alpha\right)$, we get:

$$
2 \sqrt{3} \cos ^{2} \alpha+\cos \alpha-2 \sqrt{3}=0
$$

from which

$$
\cos \alpha=\frac{-1 \pm \sqrt{1+48}}{4 \sqrt{3}}=\frac{-1 \pm 7}{4 \sqrt{3}}
$$

thus (since the root $\cos \alpha=-\frac{2}{\sqrt{3}}$ is not valid, as $|\cos \alpha|<1$)

$$
\cos \alpha=\frac{6}{4 \sqrt{3}}=\frac{\sqrt{3}}{2}
$$

From this,

$$
\alpha_{1}=30^{\circ} \pm k \cdot 360^{\circ}, \quad \alpha_{2}=330^{\circ} \pm k \cdot 360^{\circ}, \text { where } k=0,1,2 \ldots
$$

Tatai Péter (Bp. XIV., I. István g. II. o. t.)