5.19 $7+4 \sin x \cos x+1,5(\tan x+\cot x)=0$.

The translation is as follows:

5.19 $7+4 \sin x \cos x+1.5(\tan x+\cot x)=0$.
5.19 The equation is defined for $\sin x \neq 0, \cos x \neq 0$. Transform its left side:

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

We have: $2 \sin ^{2} 2 x+7 \sin 2 x+3=0 ; \sin 2 x \neq \frac{-7-5}{4}=-3$;

$$
\begin{aligned}
& \sin 2 x=\frac{-7+5}{4}=-\frac{1}{2}, 2 x=(-1)^{k+1} \cdot \frac{\pi}{6}+\pi k \\
& x=(-1)^{k+1} \cdot \frac{\pi}{12}+\frac{\pi k}{2}
\end{aligned}
$$

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