Romanian Mathematical Olympiad

From the hypothesis, it is clear that
$$\cos(\pi \log_3 (x + 6)) = \cos(\pi \log_3 (x - 2)) = \pm 1,$$
and thus, there exist $k, l \in \mathbb{Z}$, with the same parity, such that $\pi \log_3 (x + 6) = k\pi$ and $\pi \log_3 (x - 2) = l\pi$.
We get now the relations $x + 6 = 3^k$ and $x - 2 = 3^l$. By subtracting the last two equalities, we get $3^k - 3^l = 8$.
It is not difficult to see that if $k, l < 0$ the equality is not possible. Then $3^l (3^{k-l} - 1) = 8$, and thus $3^l = 1$, that is $l = 0$ and then $k = 2$. Finally we get $x = 3$.