Solve the following equation:

$$
\operatorname{tg} x+\operatorname{tg} 2 x+\operatorname{tg} 3 x+\operatorname{tg} 4 x=0
$$
I. Solution: By transforming the sum of the two extreme and two middle terms on the left side according to the identity

$$
\operatorname{tg} u+\operatorname{tg} v=\frac{\sin u}{\cos u}+\frac{\sin v}{\cos v}=\frac{\sin u \cos v+\cos u \sin v}{\cos u \cos v}=\frac{\sin (u+v)}{\cos u \cos v}
$$

our equation becomes:

$$
\begin{gathered}
(\operatorname{tg} x+\operatorname{tg} 4 x)+(\operatorname{tg} 2 x+\operatorname{tg} 3 x)=\frac{\sin 5 x}{\cos x \cos 4 x}+\frac{\sin 5 x}{\cos 2 x \cos 3 x}= \\
=\sin 5 x \frac{\cos x \cos 4 x+\cos 2 x \cos 3 x}{\cos x \cos 2 x \cos 3 x \cos 4 x}=0
\end{gathered}
$$

In the numerator, based on $\sin 2 u=2 \sin u \cos u$ and $\cos 2 u=2 \cos ^{2} u-1=1-2 \sin ^{2} u$,

$$
\begin{aligned}
\cos 3 x & =\cos (2 x+x)+\cos 2 x \cos x-\sin 2 x \sin x= \\
& =\cos x\left(\cos 2 x-2 \sin ^{2} x\right)=\cos x[\cos 2 x-(1-\cos 2 x)]= \\
& =\cos x(2 \cos 2 x-1) \\
\cos 4 x & =\cos 2 \cdot 2 x=2 \cos ^{2} 2 x-1 .
\end{aligned}
$$

Thus, simplifying by $\cos x$ (which is not zero, because $\cos x=0$ would make $\operatorname{tg} x$ undefined),

$$
\sin 5 x \frac{\cos 4 x+\cos 2 x(2 \cos 2 x-1)}{\cos 2 x \cos 3 x \cos 4 x}=\frac{\sin 5 x\left(4 \cos ^{2} 2 x-\cos 2 x-1\right)}{\cos 2 x \cos 3 x \cos 4 x}=0
$$

Therefore, $x$ can only be an angle that makes the numerator zero, i.e., it satisfies at least one of the following equations:

$$
\begin{array}{lll}
\text { a) } \sin 5 x=0 & \text { and } & \text { b) } 4 \cos ^{2} 2 x-\cos 2 x-1=0
\end{array}
$$

And every such angle is a solution to the given equation (provided it does not make the denominator zero, in which case at least one term of the equation is undefined), because we have only performed equivalent transformations.

Finally, it is sufficient to give the solutions in an interval of $180^{\circ}$, for example, the solutions between $-90^{\circ}$ and $90^{\circ}$ (excluding the endpoints, since $\operatorname{tg} x$ is undefined at these points), because the smallest period of the $\operatorname{tg} x$ function is $180^{\circ}$, and the smallest period of $\operatorname{tg} 2 x, \operatorname{tg} 3 x, \operatorname{tg} 4 x$ is half, one-third, and one-fourth of $180^{\circ}$, respectively, so $180^{\circ}$ is the period of all four terms on the left side of the equation; however, in the solution of b), we must consider that the smallest period of the cosine function is $360^{\circ}$.

Thus, the solutions of a) from $5 x=k \cdot 180^{\circ}, x=k \cdot 36^{\circ}$ are $0^{\circ}, 36^{\circ}, 72^{\circ}, 108^{\circ}, 144^{\circ}$; and the solutions of b) for which $-180^{\circ}<2 x<180^{\circ}$:

$$
\begin{gathered}
\cos 2 x=\frac{1 \pm \sqrt{17}}{8} \approx\left\{\begin{array}{l}
+0.6404 \\
-0.3904
\end{array}\right. \text {-from } \\
2 x \approx\left\{\begin{array} { l } 
{ \pm 5 0 . 1 8 ^ { \circ } , } \\
{ \pm 1 1 2 . 9 8 ^ { \circ } , }
\end{array} \quad x \approx \left\{\begin{array}{l} 
\pm 25.09^{\circ} \\
\pm 56.49^{\circ}
\end{array}\right.\right.
\end{gathered}
$$

Accordingly, the angles that can be solutions to the equation, in increasing order of absolute value, are:

$$
0^{\circ}, \quad \pm 25.09^{\circ}, \quad \pm 36^{\circ}, \quad \pm 56.49^{\circ}, \quad \pm 72^{\circ}
$$

Each of these makes all terms of the equation defined, so each is indeed a solution. This time, we only check the solutions to verify the calculations, and it is sufficient to check the values between $0^{\circ}$ and $90^{\circ}$, because if $x$ is a solution to the equation, then $-x$ is also a solution.

Gyula Veres (Pécs, Zipernovszky K. Mechanical Engineering High School III. grade)

II. Solution: Express $\operatorname{tg} 2 x, \operatorname{tg} 3 x, \operatorname{tg} 4 x$ in terms of $\operatorname{tg} x=z$:

$$
\begin{gathered}
\operatorname{tg} 2 x=\frac{2 \operatorname{tg} x}{1-\operatorname{tg}^{2} x}=\frac{2 z}{1-z^{2}} \\
\operatorname{tg} 4 x=\frac{2 \operatorname{tg} 2 x}{1-\operatorname{tg}^{2} 2 x}=\frac{\frac{4 z}{1-z^{2}}}{1-\frac{4 z^{2}}{\left(1-z^{2}\right)^{2}}}=\frac{4 z\left(1-z^{2}\right)}{1-6 z^{2}+z^{4}} \\
\operatorname{tg} 3 x=\operatorname{tg}(2 x+x)=\frac{\operatorname{tg} 2 x+\operatorname{tg} x}{1-\operatorname{tg} x \operatorname{tg} 2 x}=\frac{\frac{2 z}{1-z^{2}}+z}{1-\frac{2 z^{2}}{1-z^{2}}}=\frac{z\left(3-z^{2}\right)}{1-3 z^{2}}
\end{gathered}
$$

Thus, the equation becomes:

$$
z\left(1+\frac{2}{1-z^{2}}+\frac{3-z^{2}}{1-3 z^{2}}+\frac{4-4 z^{2}}{1-6 z^{2}+z^{4}}\right)=0
$$

According to this, one solution is $z_{1}=0$ and from this, in the open interval $\left(-90^{\circ}, 90^{\circ}\right)$, $x_{1}=0^{\circ}$, and the other solutions come from the four-term part (1). In this, the two sums (extreme and middle) have a common factor:

$$
\begin{gathered}
\frac{5-10 z^{2}+z^{4}}{1-6 z^{2}+z^{4}}+\frac{5-10 z^{2}+z^{4}}{\left(1-z^{2}\right)\left(1-3 z^{2}\right)}= \\
=\left(5-10 z^{2}+z^{4}\right)\left(\frac{1}{1-6 z^{2}+z^{4}}+\frac{1}{1-4 z^{2}+3 z^{4}}\right)=0
\end{gathered}
$$

and this makes it possible to easily calculate the solutions.

$$
z^{4}-10 z^{2}+5=0 \text {-from } z= \pm \sqrt{\frac{10 \mp \sqrt{80}}{2}}= \pm \sqrt{5 \mp 2 \sqrt{5}} \approx\left\{\begin{array}{l} 
\pm 0.7265 \\
\pm 3.078
\end{array}\right.
$$

and thus $x_{2,3} \approx \pm 36.0^{\circ}, x_{4,5} \approx 72.0^{\circ} ;$ on the other hand, from the two-term sum

$$
\begin{gathered}
\frac{2-10 z^{2}+4 z^{4}}{\left(1-6 z^{2}+z^{4}\right)\left(1-4 z^{2}+3 z^{4}\right)}=0, \text { from } 2-10 z^{2}+4 z^{4}=0 \text {-from } \\
z= \pm \sqrt{5 \mp \sqrt{17}} / 2, \quad z_{1} \approx \pm 0.4682, \quad z_{2} \approx \pm 1.510
\end{gathered}
$$

thus $x_{6,7} \approx \pm 25.09^{\circ}, x_{8,9} \approx 56.48^{\circ}$, in agreement with the first solution.