8.377. $2^{\sin ^{2} x}+4 \cdot 2^{\cos ^{2} x}=6$.

8.377. $2^{\sin ^{2} x}+4 \cdot 2^{\cos ^{2} x}=6$.
## Solution.

Let's rewrite the equation as

$$
2^{\sin ^{2} x}+4 \cdot 2^{1-\sin ^{2} x}-6=0 \Leftrightarrow\left(2^{\sin ^{2} x}\right)^{2}-6\left(2^{\sin ^{2} x}\right)+8=0
$$

Solving it as a quadratic equation in terms of $2^{\sin ^{2} x}$, we get

1) $2^{\sin ^{2} x}=2, \sin ^{2} x=1, \sin x= \pm 1, x=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z$;
2) $2^{\sin ^{2} x}=2^{2}, \sin ^{2} x=2, \varnothing$.

Answer: $x=\frac{\pi}{2}(2 k+1), k \in Z$.