8.341. $\operatorname{ctg}\left(x-25^{\circ}\right)+\operatorname{tg}\left(3 x+15^{\circ}\right)=2 \sin \left(2 x-50^{\circ}\right)$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\sin \left(x-25^{\circ}\right) \neq 0, \\ \cos \left(3 x+15^{\circ}\right) \neq 0,\end{array} \Leftrightarrow\left\{\begin{array}{l}x-25^{\circ} \neq 180^{\circ} t, \\ 3 x+15^{\circ} \neq 90^{\circ}+180^{\circ} p,\end{array} \Leftrightarrow\right.\right.$ $\Leftrightarrow\left\{\begin{array}{l}x \neq 25^{\circ}+180^{\circ} t, t \in Z, \\ x \neq 25^{\circ}+60^{\circ} p, p \in Z .\end{array}\right.$

## Write the equation in the form

$$
\begin{aligned}
& \frac{\cos \left(x-25^{\circ}\right)}{\sin \left(x-25^{\circ}\right)}+\frac{\sin \left(3 x+15^{\circ}\right)}{\cos \left(3 x+15^{\circ}\right)}-2 \sin \left(2 x-50^{\circ}\right)=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\cos \left(x-25^{\circ}\right) \cos \left(3 x+15^{\circ}\right)+\sin \left(x-25^{\circ}\right) \sin \left(3 x+15^{\circ}\right)}{\sin \left(x-25^{\circ}\right) \cos \left(3 x+15^{\circ}\right)}-2 \sin \left(2 x-50^{\circ}\right)=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\cos \left(2 x+40^{\circ}\right)}{\sin \left(x-25^{\circ}\right) \cos \left(3 x+15^{\circ}\right)}-2 \sin \left(\left(2 x+40^{\circ}\right)-90^{\circ}\right)=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\cos \left(2 x+40^{\circ}\right)}{\sin \left(x-25^{\circ}\right) \cos \left(3 x+15^{\circ}\right)}+2 \cos \left(2 x+40^{\circ}\right)=0 \Leftrightarrow \\
& \Leftrightarrow \cos \left(2 x+40^{\circ}\right)\left(1+2 \sin \left(x-25^{\circ}\right) \cos \left(3 x+15^{\circ}\right)\right)=0 .
\end{aligned}
$$

From this

1) $\cos \left(2 x+40^{\circ}\right)=0,2 x+40^{\circ}=90^{\circ}+180^{\circ} k, x_{1}=25^{\circ}+90^{\circ} k, k \in Z$;
2) $1+2 \sin \left(x-25^{\circ}\right) \cos \left(3 x+15^{\circ}\right)=0 \Leftrightarrow$

$$
\begin{aligned}
& \Leftrightarrow 1+\sin \left(x-25^{\circ}-3 x-15^{\circ}\right)+\sin \left(x-25^{\circ}+3 x+15^{\circ}\right)=0 \Leftrightarrow \\
& \Leftrightarrow 1-\sin \left(2 x+40^{\circ}\right)+\sin \left(4 x-10^{\circ}\right)=0 \Leftrightarrow \\
& \Leftrightarrow 1-\sin \left(2 x+40^{\circ}\right)+\sin \left(4 x+80^{\circ}-90^{\circ}\right)=0 \Leftrightarrow \\
& \Leftrightarrow 1-\sin \left(2 x+40^{\circ}\right)-\sin \left(90^{\circ}-\left(4 x+80^{\circ}\right)\right)=0 \Leftrightarrow \\
& \Leftrightarrow 1-\sin \left(2 x+40^{\circ}\right)-\cos \left(4 x+80^{\circ}\right)=0 \Leftrightarrow \\
& \Leftrightarrow 1-\sin \left(2 x+40^{\circ}\right)-\cos 2\left(2 x+40^{\circ}\right)=0 \Leftrightarrow \\
& \Leftrightarrow 1-\sin \left(2 x+40^{\circ}\right)-1+2 \sin ^{2}\left(2 x+40^{\circ}\right)=0 \Leftrightarrow \\
& \Leftrightarrow 2 \sin ^{2}\left(2 x+40^{\circ}\right)-\sin \left(2 x+40^{\circ}\right)=0 \Leftrightarrow \\
& \Leftrightarrow 2 \sin ^{2}\left(2 x+40^{\circ}\right)\left(2 \sin \left(2 x+40^{\circ}\right)-1\right)=0 \Leftrightarrow \\
& \Leftrightarrow\left[\begin{array} { l } 
{ \sin ( 2 x + 4 0 ^ { \circ } ) = 0 , } \\
{ \sin ( 2 x + 4 0 ^ { \circ } ) = \frac { 1 } { 2 } , }
\end{array} \Leftrightarrow \left[\begin{array}{l}
2 x+40^{\circ}=180^{\circ} n, \\
2 x+40^{\circ}=30^{\circ}+360^{\circ} \mathrm{m}, \\
2 x+40^{\circ}=150^{\circ}+360^{\circ} l,
\end{array}\right.\right. \\
& \Leftrightarrow\left[\begin{array}{l}
x_{2}=-20^{\circ}+90^{\circ} n, n \in Z \\
x_{3}=-5^{\circ}+180^{\circ} m, m \in Z \\
x_{4}=55^{\circ}+180^{\circ} l, l \in Z
\end{array}\right.
\end{aligned}
$$

Considering the domain of definition, we have $x_{1}=25^{\circ}+90^{\circ}(2 r+1)=115^{\circ}+180^{\circ} r, r \in Z$.

Answer: $x_{1}=115^{\circ}+180^{\circ} r, r \in Z ; x_{2}=-20^{\circ}+90^{\circ} n, n \in Z ; x_{3}=-5^{\circ}+180^{\circ} \mathrm{m}$, $m \in Z ; x_{4}=55^{\circ}+180^{\circ} l, l \in Z$.