Statement-1: The number of common solutions of the trigonometric 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 two.
Statement-2: The number of solutions of the equation, $$2\cos^2\theta - 3\sin\theta = 0$$ in the interval $$[0, \pi]$$ is two.

We are given two statements and need to verify their truth and whether Statement-2 explains Statement-1. Let's solve each equation step by step.

First, consider Statement-1: finding common solutions of $$2\sin^2\theta - \cos 2\theta = 0$$ and $$2\cos^2\theta - 3\sin\theta = 0$$ in $$[0, 2\pi]$$.

Start with the first equation: $$2\sin^2\theta - \cos 2\theta = 0$$. Using the identity $$\cos 2\theta = 1 - 2\sin^2\theta$$, substitute:

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

Simplify:

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

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

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

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

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

Now, find solutions in $$[0, 2\pi]$$:

• For $$\sin\theta = \frac{1}{2}$$, solutions are $$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$$.
• For $$\sin\theta = -\frac{1}{2}$$, solutions are $$\theta = \frac{7\pi}{6}, \frac{11\pi}{6}$$.

So, solutions for the first equation are $$\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$$.

Next, solve the second equation: $$2\cos^2\theta - 3\sin\theta = 0$$. Using $$\cos^2\theta = 1 - \sin^2\theta$$, substitute:

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

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

Multiply both sides by -1 and rearrange:

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

Let $$u = \sin\theta$$, so:

$$2u^2 + 3u - 2 = 0$$

Solve the quadratic equation. Discriminant $$d = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-2) = 9 + 16 = 25$$:

$$u = \frac{-3 \pm \sqrt{25}}{4} = \frac{-3 \pm 5}{4}$$

Thus,

$$u = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$$

$$u = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$$

So, $$\sin\theta = \frac{1}{2}$$ or $$\sin\theta = -2$$. Since $$\sin\theta = -2$$ is impossible (sine range is $$[-1, 1]$$), only $$\sin\theta = \frac{1}{2}$$. In $$[0, 2\pi]$$, solutions are $$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$$.

Common solutions of both equations in $$[0, 2\pi]$$ are $$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$$. There are two common solutions, so Statement-1 is true.

Now, Statement-2: number of solutions of $$2\cos^2\theta - 3\sin\theta = 0$$ in $$[0, \pi]$$. From above, the solutions are $$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$$, both in $$[0, \pi]$$ (since $$\frac{\pi}{6} \approx 0.52 < \pi \approx 3.14$$ and $$\frac{5\pi}{6} \approx 2.62 < \pi$$). Thus, there are two solutions, so Statement-2 is true.

Statement-2 states that the second equation has two solutions in $$[0, \pi]$$, but it does not mention the first equation or the full interval $$[0, 2\pi]$$. The common solutions in Statement-1 are both in $$[0, \pi]$$, but Statement-2 does not address why these are common or consider the entire interval. Therefore, Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is Option B.