We need to find the approximate values of the roots of the equation

$$
\operatorname{tg} x + \operatorname{tg} 2 x + \operatorname{tg} 3 x = 0.1
$$

Let's proceed by first determining the roots of the equation

$$
\operatorname{tg} x + \operatorname{tg} 2 x + \operatorname{tg} 3 x = 0
$$

and then, by slightly modifying these solutions, search for the approximate values of the roots of our original equation.
I. We denote the common left side of the two equations by $f(x)$. By applying the addition theorem,

$$
\operatorname{tg} 3 x=\operatorname{tg}(2 x+x)=\frac{\operatorname{tg} 2 x+\operatorname{tg} x}{1-\operatorname{tg} 2 x \cdot \operatorname{tg} x}=\frac{\frac{2 \operatorname{tg} x}{1-\operatorname{tg}^{2} x}+\operatorname{tg} x}{1-\frac{2 \operatorname{tg}^{2} x}{1-\operatorname{tg}^{2} x}}=\frac{3 \operatorname{tg} x-\operatorname{tg}^{3} x}{1-3 \operatorname{tg}^{2} x}
$$

Based on this, (2) transforms into

$$
\operatorname{tg} x+\frac{2 \operatorname{tg} x}{1-\operatorname{tg}^{2} x}+\frac{\operatorname{tg} x\left(3-\operatorname{tg}^{2} x\right)}{1-3 \operatorname{tg}^{2} x}=0
$$

It is evident that $\operatorname{tg} x=0$ is a root, i.e., $x_{1}=0$. (It is sufficient, for example, in the interval $-90^{\circ}<x<90^{\circ}$, to consider $x_{1}=0$.)

Since the ratio of the terms is approximately $1: 2: 3$, the excess of 0.0048 is expected to disappear if the first term is reduced by $0.0048 / 6=0.0008$. From $\operatorname{tg} x=0.0167$, we get $x=0.956^{\circ}=0^{\circ} 57.3^{\prime}$, and indeed $f\left(0.956^{\circ}\right)=0.1002$. Further refinement based on the tangent table is not possible.
2. $f\left(35.3^{\circ}\right)=0.708+2.840-3.511=0.037=0.1-0.063$, but $f\left(35.4^{\circ}\right)=0.711+2.872-3.442=0.1+0.041$, so an increase of $0.1^{\circ}$ results in an increase of 0.104 in $f(x)$. We need only an increase of 0.063, so we increase the first approximate value by $0.1^{\circ}$ by only $0.063 / 0.104=0.6$ of it: $x_{2}=35.36^{\circ}$.

3. According to the previous, $f\left(-35.3^{\circ}\right)=-0.037=0.1-0.137$, but with an increase of $0.2^{\circ}$, $f\left(-35.1^{\circ}\right)=-0.703-2.778+3.655=0.1+0.074$, so similarly $x_{3}=-35.17^{\circ}$.
4. From the values $f\left(60.5^{\circ}\right)=0.1+0.029$ and $f\left(60.4^{\circ}\right)=0.1+0.003$, further reduction gives $x_{4}=60.39^{\circ}$.
5. Finally, from the values $f\left(-59.6^{\circ}\right)=0.1+0.006$ and $f\left(-59.7^{\circ}\right)=0.1-0.020$, we get $x_{5}=-59.62^{\circ}$.

Thus, the approximate roots of (1) in the interval $-90^{\circ}<x<90^{\circ}$ are:

$$
-59.62^{\circ}, \quad -35.17^{\circ}, \quad 0.956^{\circ}, \quad 35.36^{\circ}, \quad 60.39^{\circ}
$$

Somogyi Árpád (Budapest, Fazekas M. gyak. g. II. o. t.)

Remark. The root near $x_{1}=0^{\circ}$ can be approximated more finely if we use the logarithm table of the tangents (lg tg) instead of the tangent table. Indeed, from the latter, based on the 4th decimal place of the mantissa, we can get back the 4 significant digits of the tangent function everywhere, and this - taking into account the characteristic $8-10=-2$ for the used $1^{\circ}, 2^{\circ}, 3^{\circ}$ - means that we can write down the 5th decimal place of the tg-values (with a maximum error of $\pm 1$). In the tangent table, however, only 3 of the 4 decimal places are significant in the interval $\left(0.6^{\circ}, 5.7^{\circ}\right)$. (This, together with the method of presenting the tangents of angles above $45^{\circ}$, is a minor inconsistency in the table.)

The calculation is as follows:

$$
\begin{array}{lr}
\lg \operatorname{tg} 1^{\circ}=8.2419-10, & \operatorname{tg} 1^{\circ}=0.01746 \\
\lg \operatorname{tg} 2^{\circ}=0.5431-2, & \operatorname{tg} 2^{\circ}=0.03492 \\
\lg \operatorname{tg} 3^{\circ}=0.7194-2, & \operatorname{tg} 3^{\circ}=0.05241
\end{array}
$$

Thus, $f\left(1^{\circ}\right)=0.1+0.00479$, and by subtracting one-sixth of the excess from $\operatorname{tg} 1^{\circ}$, we try the approximate value $\operatorname{tg} x=0.01666$. Its logarithm is $8.2217-10$, from which we get $x=0.9545^{\circ}$ (from the new table, we get $0^{\circ} 57.28^{\prime}$; we can interpolate 2 digits from both tables because the difference in the table is 45 or 76 ten-thousandths, much greater than 10). Thus, $2 x=1.909^{\circ}, 3 x=2.8635^{\circ}$, from which $\lg \operatorname{tg} 2 x=8.5228-10, \lg \operatorname{tg} 3 x=8.6991-10$, and by reversing, $\operatorname{tg} 2 x=0.03333, \operatorname{tg} 3 x=0.05001$, and finally $f\left(0.9545^{\circ}\right)=0.10000$. (However, due to the many lookups and reversals, interpolation may result in a difference in the 5th decimal place; with a 7-digit table, $\operatorname{tg} x=0.0166590$).