8.073. $\frac{\sin \left(\frac{\pi}{2}+x\right) \sin \left(\frac{\pi}{2}-x\right)}{\cos \left(\frac{\pi}{4}+x\right) \cos \left(\frac{\pi}{4}-x\right)}=3$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\cos \left(\frac{\pi}{4}+x\right) \neq 0, \\ \cos \left(\frac{\pi}{4}-x\right) \neq 0 .\end{array}\right.$

Rewrite the equation as

$$
\begin{aligned}
& \cos x \cos x=3 \cos \left(\frac{\pi}{4}+x\right) \cos \left(\frac{\pi}{4}-x\right) \\
& \cos ^{2} x-3 \cos \left(\frac{\pi}{4}+x\right) \cos \left(\frac{\pi}{4}-x\right)=0 \Leftrightarrow \\
& \Leftrightarrow \frac{1}{2}(1+\cos 2 x)-\frac{3}{2}\left(\cos \left(\frac{\pi}{4}+x-\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}+x+\frac{\pi}{4}-x\right)\right)=0 \Leftrightarrow
\end{aligned}
$$

$\Leftrightarrow 1+\cos 2 x-3(\cos 2 x+0)=0, \quad \cos 2 x=\frac{1}{2}$,

$2 x= \pm \frac{\pi}{3}+2 \pi n, \quad x= \pm \frac{\pi}{6}+\pi n, \quad n \in Z$.

Answer: $\quad x= \pm \frac{\pi}{6}+\pi n, \quad n \in Z$.