$8.440 \operatorname{tg}(\pi \cdot \operatorname{ctg} x)=\operatorname{ctg}(\pi \cdot \operatorname{tg} x)$

$8.440 \tan(\pi \cdot \cot x)=\cot(\pi \cdot \tan x)$
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-157.jpg?height=401&width=907&top_left_y=158&top_left_x=201)

If $\operatorname{tg} x=l$, then $l+\frac{1}{l}=\frac{1}{2}+k \Rightarrow \frac{1}{2}-\frac{1}{l}$ is an integer, which is only possible when $l= \pm 2$, i.e., $k=2$ or $k=-3$. Therefore, the last system is equivalent to $\operatorname{tg} x+\frac{1}{\operatorname{tg} x}=\frac{1}{2}+k, k \in Z, k \neq 2, k \neq-3 \Leftrightarrow$ $\Leftrightarrow \frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cdot \cos x}=\frac{1}{2}+k, k \in Z, k \neq 2, k \neq-3 \Leftrightarrow$ $\Leftrightarrow \sin 2 x=\frac{4}{1+2 k}, k \in Z, k \neq 2, k \neq-3$. But $\left|\frac{4}{1+2 k}\right| \leq 1 \Rightarrow k \neq 0,1,-1,-2 \Rightarrow$ $\Rightarrow 2 x=(-1)^{n} \arcsin \frac{4}{1+2 k}+\pi n,\left[\begin{array}{l}k=3,4, \ldots \\ k=-4,-5, \ldots\end{array}\right.$

Answer: $\quad x=\frac{(-1)^{n}}{2} \arcsin \frac{4}{1+2 k}+\frac{\pi n}{2},\left[\begin{array}{l}k=3,4, \ldots \\ k=-4,-5, \ldots\end{array}\right.$