8.382. $1+2^{\operatorname{tg} x}=3 \cdot 4^{-\frac{1}{\sqrt{2}} \sin \left(\frac{\pi}{4}-x\right) \cos ^{-1} x}$.
Solution.

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

From the condition we have

$$
\begin{aligned}
& 2\left(\frac{\sin \frac{\pi}{4} \cos x-\cos \frac{\pi}{4} \sin x}{\sqrt{2} \cos x}\right) \\
& 1+2^{\operatorname{tg} x}=3 \cdot 2^{\operatorname{tg} x}=2^{\frac{\cos x-\sin x}{\cos x}} \Leftrightarrow \\
& \Leftrightarrow 1+2^{\operatorname{tg} x}=3 \cdot 2^{1-\operatorname{tg} x}, 1+2^{\operatorname{tg} x}-\frac{3 \cdot 2}{2^{\operatorname{tg} x}}=0 \Leftrightarrow\left(2^{\operatorname{tg} x}\right)^{2}+2^{\operatorname{tg} x}-6=0
\end{aligned}
$$

Solving the equation as a quadratic in terms of $2^{\operatorname{tg} x}$, we have

1) $2^{\operatorname{tg} x}=-3, \varnothing$
2) $2^{\operatorname{tg} x}=2 \Leftrightarrow \operatorname{tg} x=1, x=\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k+1), k \in Z$.

Answer: $x=\frac{\pi}{4}(4 k+1), k \in Z$.