The numbers $a_1$ and $a_2$ are integers. Since $a_1 = \frac{1}{\cos x}$, $a_2 = \frac{1}{\cos(2x)}$ and $\cos(2x) = 2(\cos x)^2 - 1$, we have $a_2 = \frac{1}{2\cos^2 x - 1} = \frac{a_1^2}{2 - a_1^2}$. Now, $a_2$ is an integer, so $2 - a_1^2$ is a divisor of $a_1^2$. So, $2 - a_1^2$ divides $2 - a_1^2$ and $a_1^2$. It therefore also divides their sum, which is equal to 2. We conclude that $2 - a_1^2$ is one of the numbers $-2, -1, 1$ or $2$.
In the first case we have $a_1^2 = 4$, in the second case we have $a_1^2 = 3$, in the third case we get $a_1^2 = 1$ and in the fourth case we get $a_1^2 = 0$. Since $a_1$ is a non-zero integer, we can only have $a_1 = -2$, $a_1 = 2$, $a_1 = -1$ or $a_1 = 1$. From here it follows that $x$ can only be equal to $0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}$ or $\frac{5\pi}{3}$.
If $x = 0$, then $a_n = 1$ for all $n$. If $x = \pi$, we get the sequence $-1, 1, -1, 1, \dots$. In the remaining four cases we have $a_{n+6} = a_n$, since for all integers $a$ we have
$$\cos\left(\frac{(n+6) \cdot a\pi}{3}\right) = \cos\left(\frac{n \cdot a\pi}{3} + 2a\pi\right) = \cos\left(\frac{n \cdot a\pi}{3}\right).$$
It therefore suffices to check that the first six terms of the sequence are integers. When $x = \frac{\pi}{3}$ and $\frac{5\pi}{3}$ we have $a_1 = 2$, $a_2 = -2$, $a_3 = -1$, $a_4 = -2$, $a_5 = 2$, $a_6 = 1$. If $x = \frac{2\pi}{3}$ or $x = \frac{4\pi}{3}$, then the first six terms are $-2, -2, 1, -2, -2, 1$.
The solutions are $0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}$ and $\frac{5\pi}{3}$.