It can be seen from the given equation that $x$ must be a positive integer. Let $x = 15k + r$ where $0 \le r \le 14$ is an integer and $k$ is a nonnegative integer. Then
$$15k + r + \left[ 5k + \frac{r}{3} \right] = \left[ 10k + \frac{2r}{3} \right] + \left[ 9k + \frac{3r}{5} \right]$$
which simplifies to $k + r + \left[ \frac{r}{3} \right] = \left[ \frac{2r}{3} \right] + \left[ \frac{3r}{5} \right]$. Thus,
$$k + r + \left( \frac{r}{3} - 1 \right) < k + r + \left[ \frac{r}{3} \right] = \left[ \frac{2r}{3} \right] + \left[ \frac{3r}{5} \right] \le \frac{2r}{3} + \frac{3r}{5} = \frac{19r}{15}$$
which follows that $k < 1 - \frac{r}{15} \le 1$. But $k \ge 0$; so, we have $k = 0$.
It can now be verified that the only solutions are $x = 2, 5$.