Problem 11.1. Find all values of $x \in(-\pi, \pi)$ such that the numbers $2^{\sin x}$, $2-2^{\sin x+\cos x}$ and $2^{\cos x}$ are consecutive terms of a geometric progression.

Emil Kolev
11.1. The numbers form a geometric progression if and only if

$$
2^{\sin x} \cdot 2^{\cos x}=\left(2-2^{\sin x+\cos x}\right)^{2}
$$

This is equivalent to $4^{\sin x+\cos x}-5.2^{\sin x+\cos x}+4=0$. Setting $y=2^{\sin x+\cos x}$ gives $y^{2}-5 y+4=0$, i.e., $y=4$ or $y_{2}=1$.

If $2^{\sin x+\cos x}=4$, then $\sin x+\cos x=2$, i.e., $\sin \left(x+\frac{\pi}{4}\right)=\sqrt{2}$, which is impossible.

If $2^{\sin x+\cos x}=1$, then $\sin x+\cos x=0$, i.e., $\sin \left(x+\frac{\pi}{4}\right)=0$. Hence $x=-\frac{\pi}{4}+k \pi$, where $k$ is an integer. Since $x \in(-\pi, \pi)$, we get finally that $x=-\frac{\pi}{4}$ and $x=\frac{3 \pi}{4}$.