1. Find all real numbers $x$ from the interval $[0,2 \pi)$, for which all terms of the sequence $\mathrm{s}$ have the general term

$$
a_{n}=\frac{1}{\cos (n x)}
$$

as integers.
IV/1. The numbers $a_{1}$ and $a_{2}$ are integers. Since $a_{1}=\frac{1}{\cos x}, a_{2}=\frac{1}{\cos (2 x)}$ and $\cos (2 x)=2(\cos x)^{2}-1$, it follows that $a_{2}=\frac{1}{2 \cos ^{2} x-1}=\frac{a_{1}^{2}}{2-a_{1}^{2}}$. Since $a_{2}$ is an integer, $2-a_{1}^{2}$ must be a divisor of $a_{1}^{2}$. Therefore, $2-a_{1}^{2}$ divides both $2-a_{1}^{2}$ and $a_{1}^{2}$, and thus also their sum, which is 2. Hence, $2-a_{1}^{2}$ is one of the numbers $-2, -1, 1$, or 2.

In the first case, we get $a_{1}^{2}=4$, in the second $a_{1}^{2}=3$, in the third $a_{1}^{2}=1$, and in the last $a_{1}^{2}=0$. Since $a_{1}$ is an integer and non-zero, the possibilities are only $a_{1}=-2, a_{1}=2, a_{1}=-1$, and $a_{1}=1$. From this, it follows that the number $x$ can only be $0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}$, or $\frac{5 \pi}{3}$.

If $x=0$, then $a_{n}=1$ for every $n$. If $x=\pi$, we get the sequence $-1,1,-1,1, \ldots$ In the other four cases, we observe that $a_{n+6}=a_{n}$, because for any integer $a$,

$$
\cos \left(\frac{(n+6) \cdot a \pi}{3}\right)=\cos \left(\frac{n \cdot a \pi}{3}+2 a \pi\right)=\cos \left(\frac{n \cdot a \pi}{3}\right)
$$

Therefore, it is sufficient to check that the first six terms of the sequence are integers. In the cases $x=\frac{\pi}{3}$ and $\frac{5 \pi}{3}$, we get $a_{1}=2, a_{2}=-2, a_{3}=-1, a_{4}=-2, a_{5}=2, a_{6}=1$. If $x=\frac{2 \pi}{3}$ or $x=\frac{4 \pi}{3}$, the first six terms of the sequence are $-2,-2,1,-2,-2,1$.

The solutions to the problem are the real numbers $0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}$, and $\frac{5 \pi}{3}$.