8.002. $7+4 \sin x \cos x+1.5(\tan x+\cot x)=0$.
## Solution.

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

$$
\begin{aligned}
& 7+4 \sin x \cos x+1.5\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)=0 \Leftrightarrow \\
& \Leftrightarrow 7+4 \sin x \cos x+1.5\left(\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right)=0 \Leftrightarrow \\
& \Leftrightarrow 7+4 \sin x \cos x+\frac{1.5}{\sin x \cos x}=0 \Leftrightarrow 7+2 \sin 2 x+\frac{3}{\sin 2 x}=0
\end{aligned}
$$

From this, we obtain the equation $2 \sin ^{2} 2 x+7 \sin 2 x+3=0$, which is quadratic in $\sin 2 x$. Thus, $\sin 2 x=-3, \varnothing$, or $\sin 2 x=-\frac{1}{2}$, from which $2 x=(-1)^{k}\left(-\frac{\pi}{6}\right)+\pi k, \quad k \in Z, x=(-1)^{k+1} \frac{\pi}{12}+\frac{\pi k}{2}, \quad k \in Z$.

Answer: $\quad x=(-1)^{k+1} \frac{\pi}{12}+\frac{\pi k}{2}, \quad k \in Z$.