$8.65 \frac{\sqrt{3}}{\cos ^{2} x}<4 \tan x$.
8.65 Here the condition $\cos x \neq 0$, i.e., $x \neq \frac{\pi}{2}+\pi n$ must be satisfied. Let's perform the transformations:

$\frac{\sqrt{3}}{\cos ^{2} x}-\frac{4 \sin x}{\cos x}0$ for $x \neq \frac{\pi}{2}+\pi n$, it is sufficient to solve the inequality $\sqrt{3}-2 \sin 2 x\frac{\sqrt{3}}{2}$. By setting $z=2 x$ and plotting the graph of the function $y=\sin z$ (Fig. 8.19), we establish that $\frac{\pi}{3}+2 \pi n<2 x<\frac{2 \pi}{3}+2 \pi n$, i.e.,

$$
\begin{aligned}
& \frac{\pi}{6}+\pi n<x<\frac{\pi}{3}+\pi n . \quad \text { These intervals do not include the values } \\
& x=\frac{\pi}{2}+\pi n.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-296.jpg?height=337&width=590&top_left_y=139&top_left_x=305)

Fig. 8.19

Answer: $\left(\frac{\pi}{6}+\pi n, \frac{\pi}{3}+\pi n\right), n \in Z$.