8.395. $\left\{\begin{array}{l}9^{2 \operatorname{tg} x+\cos y}=3 \\ 9^{\cos y}-81^{\operatorname{tg} x}=2 .\end{array}\right.$
## Solution.

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

From the condition we have

$\left\{\begin{array}{l}81^{\operatorname{tg} x} \cdot 9^{\cos y}=3, \\ 9^{\cos y}-81^{\operatorname{tg} x}=2,\end{array} \Rightarrow 9^{\cos y}=2+81^{\operatorname{tg} x}, 81^{\operatorname{tg} x} \cdot\left(2+81^{\operatorname{tg} x}\right)=3\right.$,

$\left(81^{\operatorname{tg} x}\right)^{2}+2\left(81^{\operatorname{tg} x}\right)-3=0$.

Solving this equation as a quadratic equation in terms of $81^{\text {tgx }}$, we get $81^{\operatorname{tg} x}=-3, \varnothing ; 81^{\operatorname{tg} x}=1$. Hence, $\operatorname{tg} x=0, x=\pi k_{1}, k_{1} \in Z$.

Then

$$
9^{\cos y}=2+1=3,3^{2 \cos y}=3,2 \cos y=1, \cos y=\frac{1}{2}, y= \pm \frac{\pi}{3}+2 \pi k_{2}, k_{2} \in Z
$$

Answer: $x=\pi k_{1}, y= \pm \frac{\pi}{3}+2 \pi k_{2}$, where $k_{1}$ and $k_{2} \in Z$.