11.33. Solve the system of equations

$$
\left\{\begin{array}{l}
3\left(x+\frac{1}{x}\right)=4\left(y+\frac{1}{y}\right)=5\left(z+\frac{1}{z}\right) \\
x y+y z+z x=1
\end{array}\right.
$$
11.33. Answer: $(1 / 3,1 / 2,1)$ and ( $-1 / 3,-1 / 2,-1)$.

The equalities $\frac{x}{3\left(1+x^{2}\right)}=\frac{y}{4\left(1+y^{2}\right)}=\frac{z}{5\left(1+z^{2}\right)}$ show that the numbers $x, y, z$ have the same sign, and if $(x, y, z)$ is a solution to the system, then ( $-x,-y,-z$ ) is also a solution. Therefore, it is sufficient to find the positive solutions.

We use the fact that $\operatorname{tg} \varphi+\frac{1}{\operatorname{tg} \varphi}=\frac{2}{\sin 2 \varphi}$. Choose angles $\alpha$, $\beta, \gamma$, between 0 and $\pi$, such that $\operatorname{tg}(\alpha / 2)=x, \operatorname{tg}(\beta / 2)=y$, $\operatorname{tg}(\gamma / 2)=z$. Then

$$
\frac{\sin \alpha}{3}=\frac{\sin \beta}{4}=\frac{\sin \gamma}{5}
$$

and $\frac{1}{z}=\frac{x+y}{1-x y}$, i.e., $\operatorname{ctg} \frac{\gamma}{2}=\operatorname{tg} \frac{\alpha+\beta}{2}$. If $0<\alpha, \beta, \gamma<\pi$, then the equality $\operatorname{tg}\left(\frac{\pi}{2}-\frac{\gamma}{2}\right)=\operatorname{tg} \frac{\alpha+\beta}{2}$ holds only when $\frac{\pi}{2}-\frac{\gamma}{2}=\frac{\alpha+\beta}{2}$, i.e., $\alpha+\beta+\gamma=\pi$. Thus, $\alpha$, $\beta$ and $\gamma$ are the angles of a triangle with sides in the ratio $3: 4: 5$. Therefore, $\sin \alpha=\frac{3}{5}, \sin \beta=\frac{4}{5}, \gamma=\frac{\pi}{2}$. Hence, $\operatorname{tg} \frac{\alpha}{2}=\frac{1}{3}$, $\operatorname{tg} \frac{\beta}{2}=\frac{1}{2}$ and $\operatorname{tg} \frac{\gamma}{2}=1$ (one can use, for example, the formula $\operatorname{tg} \frac{\alpha}{2}=\frac{1}{\sin \alpha}-\sqrt{\frac{1}{\sin ^{2} \alpha}-1}$).