$8.441 \cos ^{-4} x-2 \cos ^{-2} x-12 \tan x-16=0$.
Solution. The equation is equivalent to the following:

$\left(1+\operatorname{tg}^{2} x\right)^{2}-2\left(1+\operatorname{tg}^{2} x\right)-12 \operatorname{tg} x-16=0 \Leftrightarrow$

$\Leftrightarrow\left(1+\operatorname{tg}^{2} x\right)^{2}-\left(2 \operatorname{tg}^{2} x+12 \operatorname{tg} x+18\right)=0 \Leftrightarrow\left(1+\operatorname{tg}^{2} x\right)^{2}-2(\operatorname{tg} x+3)^{2}=0 \Leftrightarrow$

$\Leftrightarrow\left(\operatorname{tg}^{2} x-\sqrt{2} \operatorname{tg} x+1-3 \sqrt{2}\right)\left(\operatorname{tg}^{2} x+\sqrt{2} \operatorname{tg} x+1+3 \sqrt{2}\right)=0 \Leftrightarrow$

$\Leftrightarrow \operatorname{tg}^{2} x-\sqrt{2} \operatorname{tg} x+1-3 \sqrt{2}=0 . \operatorname{tg} x=\frac{\sqrt{2} \pm \sqrt{12 \sqrt{2}-2}}{2}=\frac{1 \pm \sqrt{6 \sqrt{2}-1}}{\sqrt{2}}$.

Answer: $\quad x=\operatorname{arctg} \frac{1 \pm \sqrt{6 \sqrt{2}-1}}{\sqrt{2}}+\pi k$.