Suppose $\theta$ satisfies the equation. Then $\log(\sin(3\theta) \cdot \cos(2\theta)) = 0$. This implies $\sin(3\theta) \cdot \cos(2\theta) = 1$, so applying the Product-to-Sum Formula
$$\sin a \cdot \cos b = \frac{1}{2} (\sin(a + b) + \sin(a - b))$$
gives $\frac{1}{2}(\sin(5\theta) + \sin\theta) = 1$. Then $\sin(5\theta) + \sin\theta = 2$, so $\sin(5\theta) = \sin\theta = 1$, and the only possible solution is $\theta = \frac{\pi}{2}$.
However, this solution is not valid because $\sin(3\theta)$ is equal to $-1$, and $\log(\sin(3\theta))$ is not defined.

As in the first solution, $\sin(3\theta) \cdot \cos(2\theta) = 1$. Then either $\sin(3\theta) = 1$ and $\cos(2\theta) = 1$, or $\sin(3\theta) = -1$ and $\cos(2\theta) = -1$. However, only the first case is possible because otherwise the logarithms in the equation are not defined. In the first case, $3\theta \in \{\frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}\}$ and $2\theta \in \{0, 2\pi, 4\pi\}$, yielding $\theta \in \{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}\}$ and $\theta \in \{0, \pi, 2\pi\}$. The two sets have no common values for $\theta$, so the first case yields no solutions.