9.292 Solve the inequality: $\frac{\cos ^{2} 2 x}{\cos ^{2} x} \geq 3 \operatorname{tg} x$.
## Solution.

$\frac{\cos ^{2} 2 x}{\cos ^{2} x} \geq \frac{3 \sin x}{\cos x} \Leftrightarrow\left\{\begin{array}{l}2 \cos ^{2} 2 x \geq 3 \sin 2 x, \\ x \neq \frac{\pi}{2}+\pi k\end{array} \Leftrightarrow\left\{\begin{array}{l}2 \sin ^{2} 2 x+3 \sin 2 x-2 \leq 0, \\ x \neq \frac{\pi}{2}+\pi k\end{array} \Leftrightarrow\right.\right.$

$\Leftrightarrow\left\{\begin{array}{l}-2 \leq \sin 2 x \leq \frac{1}{2}, \\ x \neq \frac{\pi}{2}+\pi k\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin x \leq \frac{1}{2}, \\ x \neq \frac{\pi}{2}+\pi k\end{array} \Leftrightarrow\left\{\begin{array}{l}-\frac{7 \pi}{6}+2 \pi n \leq 2 x \leq \frac{\pi}{6}+2 \pi n, \\ x \neq \frac{\pi}{2}+\pi k\end{array} \Leftrightarrow\right.\right.\right.$

$\Leftrightarrow\left\{\begin{array}{l}-\frac{7 \pi}{12}+\pi n \leq x \leq \frac{\pi}{12}+\pi n \\ x \neq-\frac{\pi}{2}+\pi n\end{array}\right.$

Answer: $\quad x \in\left[-\frac{7 \pi}{12}+\pi n ;-\frac{\pi}{2}+\pi n\right) \cup\left(-\frac{\pi}{2}+\pi n ; \frac{\pi}{12}+\pi n\right]$.