8.402. $\left\{\begin{array}{l}2^{\cos x}+2^{\cos ^{-1} y}=5, \\ 2^{\cos x} \cdot 2^{\cos ^{-1} y}=4 .\end{array}\right.$

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8.402. $\left\{\begin{array}{l}2^{\cos x}+2^{\cos ^{-1} y}=5, \\ 2^{\cos x} \cdot 2^{\cos ^{-1} y}=4 .\end{array}\right.$
## Solution.

From the condition we have

$$
2^{\cos ^{-1} y}=5-2^{\cos x}, 2^{\cos x} \cdot\left(5-2^{\cos x}\right)=4,\left(2^{\cos x}\right)^{2}-5\left(2^{\cos x}\right)+4=0
$$

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

$$
2^{\cos x}=1, \cos x=0, x=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z
$$

or

$$
2^{\cos x}=2^{2}, \cos x=2, \varnothing
$$

Substituting the value $2^{\cos x}=1$ into the first equation of the system, we find $2^{\cos ^{-1} y}=4, \cos ^{-1} y=2, \cos y=\frac{1}{2}, y= \pm \frac{\pi}{3}+2 \pi k_{1}=\frac{\pi}{3}\left(6 k_{1} \pm 1\right), k_{1} \in Z$.

Answer: $x=\frac{\pi}{2}(2 k+1), y=\frac{\pi}{3}\left(6 k_{1} \pm 1\right)$, where $k$ and $k_{1} \in Z$.