Let's find all the numbers $\alpha$ for which the

$$
\cos \alpha, \quad \cos 2 \alpha, \quad \cos 2^{2} \alpha, \quad \ldots, \quad \cos 2^{n} \alpha, \quad \ldots
$$

infinite sequence has every element negative.
Solution. The cosine function is even and periodic with $2 \pi$, so if an $\alpha$ satisfies the problem's requirement, then $\pm \alpha+2 k \pi$ also satisfies it for any integer $k$. Therefore, it is sufficient to find such $\alpha$ numbers that meet the problem's condition and $0\pi / 2$, and due to the $\cos 2 \alpha<0$ condition, $\alpha \neq \pi$. Thus,

$$
\frac{\pi}{2}<\alpha<\pi
$$

The $\cos 2^{n} \alpha<0$ condition generally means that there is an integer $l_{n}$ such that $\pi / 2+2 l_{n} \pi<2^{n} \alpha<3 \pi / 2+2 l_{2} \pi$, or with the notation $\alpha_{0}=2 \alpha / \pi$

$$
4 l_{n}+1<2^{n} \alpha_{0}<4 l_{n}+3
$$

Condition (1) implies that $\left[\alpha_{0}\right]=1$. We will prove that in the base-4 number system, $\alpha_{0}=1,=1,111 \ldots$, that is, every digit of $\alpha_{0}$ in the base-4 number system is 1.

Condition (2) precisely means that when the number $2^{n} \alpha_{0}$ is written in the base-4 number system, a 1 or 2 stands in the units place. But by multiplying by an appropriate power of four, any digit of the fractional part of $\alpha_{0}$ can be moved to the units place. (The $m$-th digit is moved to the units place by multiplying by $4^{m}=2^{2 m}$.) Therefore, every digit of the fractional part of $\alpha_{0}$ is 1 or 2.

But the same is true for $2 \alpha_{0}$ as well, since any digit of its fractional part can also be moved to the units place by multiplying by an appropriate power of two. Suppose there is a 2 in $\alpha_{0}$ somewhere. Let the first 2 be the $m$-th digit after the decimal point: $\alpha_{0}=1,1 \ldots 112 \ldots$ ( $m$ 1's). Then $4^{m-1} \alpha_{0}=11 \ldots 11,2 \ldots$, and multiplying this by two, a 3 will appear in the units place, since the fractional part is greater than $1 / 2$, and thus there is a carry. Thus $2^{2 m-1} \alpha_{0}=22 \ldots 23,1 \ldots$ or $22 \ldots 23,0 \ldots$, which contradicts the fact that a 1 or 2 stands in the units place of $2^{2 m-1} \alpha_{0}$. This proves that $\alpha_{0}=1, \dot{1}=4 / 3$, or $\alpha=2 \pi / 3$. Therefore, the numbers of the form $\pm 2 \pi / 3+2 k \pi(k=0, \pm 1, \pm 2, \ldots)$ are the only ones that can satisfy the problem's condition, and they do satisfy it.

Remark. The problem and its solution can be found in the following book: Pataki J.-Reiman I.: Mathematical Competitions $1989,45-47$. pages.