If the sum of solutions of the system of equations $$2\sin^2\theta - \cos 2\theta = 0$$ and $$2\cos^2\theta + 3\sin\theta = 0$$ in the interval $$[0, 2\pi]$$ is $$k\pi$$, then $$k$$ is equal to ______.

We need to find the sum of common solutions of $$2\sin^2\theta - \cos 2\theta = 0$$ and $$2\cos^2\theta + 3\sin\theta = 0$$ in $$[0, 2\pi]$$.

$$2\sin^2\theta - \cos 2\theta = 0$$

Using $$\cos 2\theta = 1 - 2\sin^2\theta$$:

$$2\sin^2\theta - (1 - 2\sin^2\theta) = 0$$

$$4\sin^2\theta - 1 = 0$$

$$\sin^2\theta = \dfrac{1}{4}$$

$$\sin\theta = \pm\dfrac{1}{2}$$

Solutions in $$[0, 2\pi]$$: $$\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$$

$$2\cos^2\theta + 3\sin\theta = 0$$

Using $$\cos^2\theta = 1 - \sin^2\theta$$:

$$2(1 - \sin^2\theta) + 3\sin\theta = 0$$

$$2 - 2\sin^2\theta + 3\sin\theta = 0$$

$$2\sin^2\theta - 3\sin\theta - 2 = 0$$

$$(2\sin\theta + 1)(\sin\theta - 2) = 0$$

Since $$\sin\theta \neq 2$$: $$\sin\theta = -\dfrac{1}{2}$$

Solutions in $$[0, 2\pi]$$: $$\theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$$

Common solutions: $$\theta = \dfrac{7\pi}{6}$$ and $$\theta = \dfrac{11\pi}{6}$$

$$\dfrac{7\pi}{6} + \dfrac{11\pi}{6} = \dfrac{18\pi}{6} = 3\pi$$

So $$k\pi = 3\pi$$, giving $$k = 3$$.

The correct answer is $$\boxed{3}$$.