Let's calculate the acute and obtuse angles that satisfy the following equation:

$$
\cot \alpha \cdot \tan 2 \alpha-2=\tan \alpha \cdot \cot 2 \alpha
$$
$$
\begin{gathered}
\frac{1}{\tan \alpha} \cdot \frac{2 \tan \alpha}{1-\tan ^{2} \alpha}-2=\tan \alpha \cdot \frac{1-\tan ^{2} \alpha}{2 \tan \alpha} \\
\frac{2}{1-\tan ^{2} \alpha}-\frac{1-\tan ^{2} \alpha}{2}=2
\end{gathered}
$$

This equation can also be written as:

$$
\left(\frac{2}{1-\tan ^{2} \alpha}\right)^{2}-2\left(\frac{2}{1-\tan ^{2} \alpha}\right)-1=0
$$

from which

$$
\begin{gathered}
\frac{2}{1-\tan ^{2} \alpha}=1 \pm \sqrt{2} \\
\tan ^{2} \alpha=1-\frac{2}{1 \pm \sqrt{2}}=\frac{ \pm \sqrt{2}-1}{1 \pm \sqrt{2}}=( \pm \sqrt{2}-1)^{2}
\end{gathered}
$$

and thus

$$
\tan \alpha= \pm( \pm \sqrt{2}-1)
$$

From this, the values of $\alpha$ are:

$$
\alpha_{1}=22^{\circ} 30^{\prime}, \quad \alpha_{2}=67^{\circ} 30^{\prime}, \quad \alpha_{3}=112^{\circ} 30^{\prime}, \quad \alpha_{4}=157^{\circ} 30^{\prime}
$$

The problem was solved by: Friedmann Bernát, S.-A.-Ujhely; Goldberger Leó, Grünhut Béla and Visnya Aladár, Pécs; Hofbauer Ervin, Kántor Nándor, Langheim Pál István and Pósch Gyula, Budapest; Schiller Jenő and Szabó Gusztáv, Győr; Suschnik József, Kecskemét; Szabó István főr. VI. o. t. Debreczen.