$8.434 \cos ^{-4} x+8 \cos ^{-1} x-7=0$.

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$8.434 \cos ^{-4} x+8 \cos ^{-1} x-7=0$.
Solution. Let $y=\frac{1}{\cos x}$, then $y^{4}+8 y-7=0 \Leftrightarrow$

$$
\begin{aligned}
& y^{4}+2 y^{2}+1-2 y^{2}+8 y-8=0 \Leftrightarrow\left(y^{2}+1\right)^{2}-2(y-2)^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(y^{2}+y \sqrt{2}-2 \sqrt{2}+1\right)\left(y^{2}-y \sqrt{2}+2 \sqrt{2}+1\right)=0 \Leftrightarrow y^{2}+y \sqrt{2}+(1-2 \sqrt{2})=0 \Leftrightarrow \\
& \Leftrightarrow y=\frac{-\sqrt{2} \pm \sqrt{8 \sqrt{2}-2}}{2} \text { But } \frac{2}{-\sqrt{2}+\sqrt{8 \sqrt{2}-2}}>1 . \text { Therefore }
\end{aligned}
$$

$$
\cos x=\frac{2}{-\sqrt{2}-\sqrt{8 \sqrt{2}-2}}, x= \pm \arccos \frac{-2}{\sqrt{2}+\sqrt{8 \sqrt{2}-2}}+2 \pi k
$$

Answer: $\quad x= \pm \arccos \frac{-2}{\sqrt{2}+\sqrt{8 \sqrt{2}-2}}+2 \pi k$