The number of solutions of $$\sin 3x = \cos 2x$$, in the interval $$\left(\frac{\pi}{2}, \pi\right)$$ is:

We need to find the number of solutions of $$\sin 3x = \cos 2x$$ in the interval $$\left(\frac{\pi}{2}, \pi\right)$$.

We rewrite $$\cos 2x$$ as $$\sin\left(\frac{\pi}{2} - 2x\right)$$, so the equation becomes:

$$\sin 3x = \sin\left(\frac{\pi}{2} - 2x\right)$$

The general solution of $$\sin A = \sin B$$ is $$A = n\pi + (-1)^n B$$, where $$n$$ is an integer.

Case 1: $$3x = 2k\pi + \left(\frac{\pi}{2} - 2x\right)$$ (when $$n$$ is even):

$$5x = 2k\pi + \frac{\pi}{2}$$

$$x = \frac{(4k + 1)\pi}{10}$$

For $$x \in \left(\frac{\pi}{2}, \pi\right)$$: we need $$\frac{\pi}{2} < \frac{(4k+1)\pi}{10} < \pi$$, which gives $$5 < 4k + 1 < 10$$, so $$4 < 4k < 9$$, meaning $$1 < k < 2.25$$.

The only integer value is $$k = 2$$, giving $$x = \frac{9\pi}{10}$$. We verify: $$\frac{\pi}{2} < \frac{9\pi}{10} < \pi$$. This is valid.

Case 2: $$3x = (2k+1)\pi - \left(\frac{\pi}{2} - 2x\right)$$ (when $$n$$ is odd):

$$3x = (2k+1)\pi - \frac{\pi}{2} + 2x$$

$$x = \frac{(4k+1)\pi}{2}$$

For $$x \in \left(\frac{\pi}{2}, \pi\right)$$: we need $$\frac{\pi}{2} < \frac{(4k+1)\pi}{2} < \pi$$, which gives $$1 < 4k + 1 < 2$$, so $$0 < 4k < 1$$, meaning $$0 < k < 0.25$$.

No integer value of $$k$$ satisfies this. So there are no solutions from this case.

The total number of solutions is 1.

The answer is Option D: 1.