Solve the following equation:

$\tan x + \tan 2x + \tan 3x + \tan 4x = 0$.
The numbers $x, 2x, 3x, 4x$ cannot be equal to any odd multiple of $\pi / 2$, otherwise the left side of (1) would be undefined. On the other hand, we recognize that for every integer $k$, $x = k\pi$ satisfies the equation, which we exclude in further searching for roots. Together with every root $x_0$, $x_0 + k\pi$, as well as $-x_0$ and $-x_0 + k\pi$ are also roots, so it suffices to restrict ourselves to the open interval $(0, \pi / 2)$.

Pairing the extreme and intermediate terms of (1),

\[
\begin{aligned}
\operatorname{tg} x + \operatorname{tg} 4x & = \frac{\sin x}{\cos x} + \frac{\sin 4x}{\cos 4x} = \frac{\sin (x + 4x)}{\cos x \cos 4x} \\
\operatorname{tg} 2x + \operatorname{tg} 3x & = \frac{\sin 5x}{\cos 2x \cos 3x}
\end{aligned}
\]

Substituting these into (1) and using the identity $2 \cos u \cos v = \cos (u + v) + \cos (u - v)$,

\[
\begin{aligned}
& \sin 5x \left( \frac{1}{\cos x \cos 4x} + \frac{1}{\cos 2x \cos 3x} \right) = \\
& = \frac{\sin 5x (2 \cos 5x + \cos 3x + \cos x)}{2 \cos x \cos 4x \cos 2x \cos 3x} = 0
\end{aligned}
\]

This is satisfied if

\[
\begin{gathered}
\sin 5x = 0, \text{ and if } \\
2 \cos 5x + \cos 3x + \cos x = 0
\end{gathered}
\]

The solution to the first equation in the specified interval is

\[
5x = k\pi, \quad x = k \frac{\pi}{5}, \quad k = 1, 2
\]

Let the left side of (2) be denoted by $y$. This can be transformed using the identity $2 \sin u \cos v = \sin (u + v) - \sin (v - u)$ and other known identities as follows:

\[
\begin{aligned}
y & = \frac{2y \sin x}{2 \sin x} = \frac{1}{2 \sin x} \{ 2 (\sin 6x - \sin 4x) + (\sin 4x - \sin 2x) + \sin 2x \} = \\
& = \frac{2 \sin 6x - \sin 4x}{2 \sin x} = \frac{2 \sin (4x + 2x) - \sin 4x}{2 \sin x} = \\
& = \frac{2 \cos 4x \sin 2x - \sin 4x (1 - 2 \cos 2x)}{2 \sin x} = \\
& = \frac{\sin 2x}{\sin x} \{ \cos 4x - \cos 2x (1 - 2 \cos 2x) \} = \\
& = 2 \cos x (4 \cos^2 2x - \cos 2x - 1)
\end{aligned}
\]

Accordingly, considering the above, additional roots can only be given by the equation

\[
4 \cos^2 2x - \cos 2x - 1 = 0
\]

From this, in degrees, in the interval $(0, \pi)$,

\[
\begin{array}{rlrl}
\cos 2x & = \frac{1 + \sqrt{17}}{8} = 0.6404 \quad \text{and} \quad \cos 2x & = \frac{1 - \sqrt{17}}{8} = -0.3904, \\
2x & = 50^\circ 11', & 2x & = 112^\circ 59'.
\end{array}
\]

Finally, collecting the found roots, (1) is satisfied in the interval $0^\circ, 180^\circ$ by the following values:

\[
\begin{gathered}
x_1 = 0^\circ; \quad x_2 = 36^\circ, \quad x_3 = 72^\circ, \quad x_4 = 108^\circ, \quad x_5 = 144^\circ \\
x_6 = 25^\circ 5', \quad x_7 = 154^\circ 55', \quad x_8 = 56^\circ 29', \quad x_9 = 123^\circ 31'
\end{gathered}
\]

Kurucz Erzsébet (Polgár, József A. Gym., III. o. t.)