If $$m$$ and $$n$$ respectively are the numbers of positive and negative value of $$\theta$$ in the interval $$[-\pi, \pi]$$ that satisfy the equation $$\cos 2\theta \cos \frac{\theta}{2} = \cos 3\theta \cos \frac{9\theta}{2}$$, then $$mn$$ is equal to _____.

We need to find $$mn$$ where $$m$$ and $$n$$ are the numbers of positive and negative values of $$\theta$$ in $$[-\pi, \pi]$$ satisfying $$\cos 2\theta \cos \frac{\theta}{2} = \cos 3\theta \cos \frac{9\theta}{2}$$.

Apply the product-to-sum formula.

Using $$2\cos A \cos B = \cos(A-B) + \cos(A+B)$$:

LHS: $$\cos 2\theta \cos \frac{\theta}{2} = \frac{1}{2}\left[\cos\frac{3\theta}{2} + \cos\frac{5\theta}{2}\right]$$

RHS: $$\cos 3\theta \cos \frac{9\theta}{2} = \frac{1}{2}\left[\cos\left(-\frac{3\theta}{2}\right) + \cos\frac{15\theta}{2}\right] = \frac{1}{2}\left[\cos\frac{3\theta}{2} + \cos\frac{15\theta}{2}\right]$$

Simplify.

Setting LHS = RHS:

$$ \cos\frac{5\theta}{2} = \cos\frac{15\theta}{2} $$

This gives: $$\cos\frac{15\theta}{2} - \cos\frac{5\theta}{2} = 0$$

Using $$\cos C - \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2}$$:

$$ -2\sin(5\theta)\sin\frac{5\theta}{2} = 0 $$

Solve each factor.

Case 1: $$\sin(5\theta) = 0 \implies 5\theta = n\pi \implies \theta = \frac{n\pi}{5}$$

In $$[-\pi, \pi]$$: $$n = -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5$$

Case 2: $$\sin\frac{5\theta}{2} = 0 \implies \frac{5\theta}{2} = n\pi \implies \theta = \frac{2n\pi}{5}$$

In $$[-\pi, \pi]$$: $$n = -2, -1, 0, 1, 2$$, giving $$\theta = -\frac{4\pi}{5}, -\frac{2\pi}{5}, 0, \frac{2\pi}{5}, \frac{4\pi}{5}$$

All of these are already included in Case 1.

Count positive and negative solutions.

The distinct solutions are $$\theta = \frac{n\pi}{5}$$ for $$n = -5, -4, ..., 4, 5$$.

Positive values ($$\theta > 0$$): $$\frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}, \pi$$ → $$m = 5$$

Negative values ($$\theta < 0$$): $$-\frac{\pi}{5}, -\frac{2\pi}{5}, -\frac{3\pi}{5}, -\frac{4\pi}{5}, -\pi$$ → $$n = 5$$

Therefore: $$mn = 5 \times 5 = \boxed{25}$$.