The given equation is $2 \sin x \sin 3x \sin 5x + \sin 5x \cos 4x = 0$.
Factor out the common term $\sin 5x$:
$\sin 5x (2 \sin x \sin 3x + \cos 4x) = 0$.
This gives two possibilities for the solutions:
Case 1: $\sin 5x = 0$.
This implies $5x = n\pi$ for any integer $n$. So, $x = \frac{n\pi}{5}$.
We are given the interval $x \in (-\pi, \pi)$, so $-\pi<\frac{n\pi}{5}<\pi$.
Dividing by $\pi$ gives $-1<\frac{n}{5}<1$, which means $-5<n<5$.
The possible integer values for $n$ are $-4, -3, -2, -1, 0, 1, 2, 3, 4$. This gives 9 distinct solutions.
Case 2: $2 \sin x \sin 3x + \cos 4x = 0$.
Using the product-to-sum formula $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$, we have:
$2 \sin 3x \sin x = \cos(3x-x) - \cos(3x+x) = \cos(2x) - \cos(4x)$.
Substituting this into the equation for Case 2:
$(\cos 2x - \cos 4x) + \cos 4x = 0$.
$\cos 2x = 0$.
This implies $2x = (2k+1)\frac{\pi}{2}$ for any integer $k$. So, $x = (2k+1)\frac{\pi}{4}$.
For the interval $x \in (-\pi, \pi)$, we have $-\pi<(2k+1)\frac{\pi}{4}<\pi$.
Multiplying by $4/\pi$ gives $-4<2k+1<4$, which means $-5<2k<3$, or $-2.5<k<1.5$.
The possible integer values for $k$ are $-2, -1, 0, 1$. This gives 4 distinct solutions: $-\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}$.
We must check if there is any overlap between the solutions from Case 1 and Case 2. An overlap would occur if $\frac{n}{5} = \frac{2k+1}{4}$ for some allowed integers $n, k$. This gives $4n = 5(2k+1)$. The left side is even, while the right side is odd. This is impossible, so there are no common solutions.
Total number of solutions = (Solutions from Case 1) + (Solutions from Case 2) = $9 + 4 = 13$.