67th 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 $x=3$.