8.335. $\operatorname{tg}\left(35^{\circ}+x\right) \operatorname{ctg}\left(10^{\circ}-x\right)=. \frac{2}{3}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\cos \left(35^{\circ}+x\right) \neq 0 \\ \sin \left(10^{\circ}-x\right) \neq 0\end{array}\right.$

Rewrite the equation as

$$
\begin{aligned}
& \operatorname{tg}\left(35^{\circ}+x\right) \operatorname{ctg}\left(90^{\circ}-\left(80^{\circ}+x\right)\right)=\frac{2}{3} \Leftrightarrow \\
& \Leftrightarrow \operatorname{tg}\left(35^{\circ}+x\right) \operatorname{tg}\left(45^{\circ}+\left(35^{\circ}+x\right)\right)=\frac{2}{3} \Leftrightarrow
\end{aligned}
$$

$$
\begin{aligned}
& \Leftrightarrow \operatorname{tg}\left(35^{\circ}+x\right) \cdot \frac{\operatorname{tg} 45^{\circ}+\operatorname{tg}\left(35^{\circ}+x\right)}{1-\operatorname{tg} 45^{\circ} \operatorname{tg}\left(35^{\circ}+x\right)}=\frac{2}{3} \Leftrightarrow \\
& \Leftrightarrow \operatorname{tg}\left(35^{\circ}+x\right) \cdot \frac{1+\operatorname{tg}\left(35^{\circ}+x\right)}{1-\operatorname{tg}\left(35^{\circ}+x\right)}=\frac{2}{3} \Leftrightarrow 3 \operatorname{tg}^{2}\left(35^{\circ}+x\right)+5 \operatorname{tg}\left(35^{\circ}+x\right)-2=0
\end{aligned}
$$

Solving this equation as a quadratic equation in $\operatorname{tg}\left(35^{\circ}+x\right)$, we get

$$
\operatorname{tg}\left(35^{\circ}+x\right)=\frac{1}{3}, 35^{\circ}+x=\operatorname{arctg} \frac{1}{3}+180^{\circ} k, x_{1}=\operatorname{arctg} \frac{1}{3}-35^{\circ}+180^{\circ} k, k \in Z
$$

or

$$
\operatorname{tg}\left(35^{\circ}+x\right)=-2,35^{\circ}+x=\operatorname{arctg}(-2)+180^{\circ} n, \quad x_{2}=-\operatorname{arctg} 2-35^{\circ}+180^{\circ} n
$$

$n \in Z$.

Answer: $x_{1}=\operatorname{arctg} \frac{1}{3}-35^{\circ}+180^{\circ} k, k \in Z, x_{2}=-\operatorname{arctg} 2-35^{\circ}+180^{\circ} n, n \in Z$. 8.336. $2 \operatorname{tg} x+\operatorname{tg} 2 x+2 \operatorname{tg} x \operatorname{tg} 3 x+\operatorname{tg} 2 x \operatorname{tg} 3 x=0$.

## Solution.

Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos 2 x \neq 0, \\ \cos 3 x \neq 0 .\end{array}\right.$

Rewrite the equation as

$$
\begin{aligned}
& (2 \operatorname{tg} x+2 \operatorname{tg} x \operatorname{tg} 3 x)+(\operatorname{tg} 2 x+\operatorname{tg} 2 x \operatorname{tg} 3 x)=0 \Leftrightarrow \\
& \Leftrightarrow 2 \operatorname{tg} x(1+\operatorname{tg} 3 x)+\operatorname{tg} 2 x(1+\operatorname{tg} 3 x)=0,(1+\operatorname{tg} 3 x)(2 \operatorname{tg} x+\operatorname{tg} 2 x)=0
\end{aligned}
$$

From this,

1) $1+\operatorname{tg} 3 x=0, \operatorname{tg} 3 x=-1,3 x=-\frac{\pi}{4}+\pi k, x_{1}=-\frac{\pi}{12}+\frac{\pi k}{3}, k \in Z$;
2) $2 \operatorname{tg} x+\operatorname{tg} 2 x=0 \Leftrightarrow 2 \operatorname{tg} x+\frac{2 \operatorname{tg} x}{1-\operatorname{tg}^{2} x}=0,2 \operatorname{tg} x\left(1+\frac{1}{1-\operatorname{tg}^{2} x}\right)=0$.

From this, either $\operatorname{tg} x=0, x_{2}=\pi n, n \in Z$, or $1+\frac{1}{1-\operatorname{tg}^{2} x}=0, \operatorname{tg} x= \pm \sqrt{2}$, $x_{3}= \pm \operatorname{arctg} \sqrt{2}+\pi l, l \in Z$. Considering the domain of definition, $x_{1}=-\frac{\pi}{12}+\frac{\pi k}{3}, k \neq 3 m+1$, $m \in Z$.

Answer: $x_{1}=-\frac{\pi}{12}+\frac{\pi k}{3}, k \neq 3 m+1 ; x_{2}=\pi n, x_{3}= \pm \operatorname{arctg} \sqrt{2}+\pi l$, where $k, l$ and $m \in Z$.