First, rewrite the equation as $3^3 \cdot 3^{3\sin x} = (3^2)^{\cos^2 x}$ and then
$$3^{3+3\sin x} = 3^{2\cos^2 x}.$$
This implies $\log_3(3^{3+3\sin x}) = \log_3(3^{2\cos^2 x})$ or $3 + 3\sin x = 2\cos^2 x$. Since $\cos^2 x = 1 - \sin^2 x$ we have
$$1 + 3\sin x + 2\sin^2 x = 0$$
and
$$(1 + \sin x)(1 + 2\sin x) = 0.$$
This implies that $\sin x = -1$ or $\sin x = -1/2$. The only real number $x$ from the interval $[0, 2\pi)$ satisfying the first condition is $x = \frac{3\pi}{2}$. There are two real numbers $x \in [0, 2\pi)$, such that $\sin x = -1/2$, namely $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$.

The solutions are $x = \frac{3\pi}{2}$, $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$.