8.075. $9^{\cos x}=9^{\sin x} \cdot 3^{\frac{2}{\cos x}}$.

Domain of definition: $\cos x \neq 0$.
## Solution.

From the condition

$$
3^{2 \cos x}=3^{2 \sin x} \cdot 3^{\frac{2}{\cos x}} \Leftrightarrow 3^{2 \cos x}=3^{2 \sin x+\frac{2}{\cos x}} \Leftrightarrow 2 \cos x=2 \sin x+\frac{2}{\cos x}
$$

From this,

$\cos ^{2} x-\sin x \cos x-1=0, \sin x \cos x+1-\cos ^{2} x=0$, $\sin x \cos x+\sin ^{2} x=0, \sin x(\cos x+\sin x)=0$.

Then:

1) $\sin x=0, x_{1}=\pi n, n \in Z$
2) $\cos x+\sin x=0 \Leftrightarrow \operatorname{tg} x=1, x_{2}=-\frac{\pi}{4}+\pi k, k \in Z$.

Answer: $\quad x_{1}=\pi n ; x_{2}=-\frac{\pi}{4}+\pi k, n, k \in Z$.