6. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& \operatorname{tg}^{2} x+2 \operatorname{cotg}^{2} 2 y=1, \\
& \operatorname{tg}^{2} y+2 \operatorname{cotg}^{2} 2 z=1, \\
& \operatorname{tg}^{2} z+2 \operatorname{cotg}^{2} 2 x=1
\end{aligned}
$$

(J. Švrček, P. Calábek)
Solution. For each admissible $\varphi$ we have

$$
2 \operatorname{cotg}^{2} 2 \varphi=2\left(\frac{\cos ^{2} \varphi-\sin ^{2} \varphi}{2 \sin \varphi \cos \varphi}\right)^{2}=\frac{1}{2}\left(\operatorname{tg}^{2} \varphi+\operatorname{cotg}^{2} \varphi-2\right) .
$$

Let $\operatorname{tg}^{2} x=a, \operatorname{tg}^{2} y=b \operatorname{atg}^{2} z=c$, where $a, b, c$ are positive real numbers. The given system can thus be transformed into the form

$$
\begin{aligned}
& a+\frac{1}{2}\left(b+\frac{1}{b}\right)=2, \\
& b+\frac{1}{2}\left(c+\frac{1}{c}\right)=2, \\
& c+\frac{1}{2}\left(a+\frac{1}{a}\right)=2 .
\end{aligned}
$$

Without loss of generality, assume that $a \geqq b \geqq c$. Under this ordering, it follows from the previous system of equations that

$$
b+\frac{1}{b} \leqq c+\frac{1}{c} \leqq a+\frac{1}{a}
$$

Since for every positive $x$ we have $x+1 / x \geqq 2$, it follows from the system (1) that $0<a, b, c \leqq 1$. The function $f(x)=x+1 / x$ is, however, decreasing on the interval $(0 ; 1\rangle$, so the inequality also holds

$$
a+\frac{1}{a} \leqq b+\frac{1}{b} \leqq c+\frac{1}{c}
$$

This, together with the previous inequalities, gives $a=b=c$.

It remains to determine all $u \in(0 ; 1\rangle$ that satisfy the equation

$$
u+\frac{1}{2}\left(u+\frac{1}{u}\right)=2
$$

After a simple rearrangement, we obtain the quadratic equation

$$
3 u^{2}-4 u+1=0, \quad \text { i.e. } \quad(u-1)(3 u-1)=0 .
$$

This quadratic equation has exactly two positive real roots $u_{1}=1$ and $u_{2}=\frac{1}{3}$. Considering the substitutions used and the periodicity of the tangent function, the solutions to the given system of equations are precisely the following triples $(x, y, z)$ of real numbers

$$
\left(\frac{\pi}{4}+k_{1} \frac{\pi}{2}, \frac{\pi}{4}+k_{2} \frac{\pi}{2}, \frac{\pi}{4}+k_{3} \frac{\pi}{2}\right) \text { and }\left( \pm \frac{\pi}{6}+k_{1} \pi, \pm \frac{\pi}{6}+k_{2} \pi, \pm \frac{\pi}{6}+k_{3} \pi\right) \text {, }
$$

where $k_{1}, k_{2}, k_{3}$ are any integers and the three signs in the triple of the second type are chosen arbitrarily, i.e., independently of each other.