Problem 569

## Solution.

Domain of definition: $\left\{\begin{array}{l}\cos (x+1) \neq 0, \\ \sin (2 x+3) \neq 0 .\end{array}\right.$

From the condition we have

$$\begin{aligned} & \operatorname{tg}(x+1)=\operatorname{tg}(2 x+3) \Leftrightarrow \operatorname{tg}(x+1)-\operatorname{tg}(2 x+3)=0 \Leftrightarrow \\ & \Leftrightarrow \frac{-\sin (x+2)}{\cos (x+1) \cos (2 x+3)}=0 \Rightarrow \sin (x+2)=0, x+2=\pi k \\ & x=-2+\pi k, k \in Z \end{aligned}$$

Answer: $x=-2+\pi k, k \in Z$.