8.281. $2 \sin 2 x+3 \tan x=5$.
## Solution.

From the condition we have

$$
\begin{aligned}
& \frac{4 \tan x}{1+\tan^{2} x}+3 \tan x-5=0 \Leftrightarrow 3 \tan^{3} x-5 \tan^{2} x+7 \tan x-5=0 \\
& 3 \tan^{3} x-3 \tan^{2} x-2 \tan^{2} x+2 \tan x+5 \tan x-5=0 \\
& 3 \tan^{2} x(\tan x-1)-2 \tan x(\tan x-1)+5(\tan x-1)=0 \\
& (\tan x-1)\left(3 \tan^{2} x-2 \tan x+5\right)=0 \Rightarrow \tan x=1, x_{1}=\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k+1) \\
& k \in \mathbb{Z} ; 3 \tan^{2} x-2 \tan x+5 \neq 0(D<0), \varnothing \\
& \text { Answer: } x=\frac{\pi}{4}(4 k+1), k \in \mathbb{Z}
\end{aligned}
$$