The sum of all values of $$\theta \in [0,2\pi]$$ satisfying $$2\sin^{2}\theta =\cos2\theta \text{ and }2\cos^{2}\theta =3\sin\theta$$ is

We are given the system of equations:

$$2\sin^{2}\theta = \cos 2\theta \quad \text{(1)}$$

$$2\cos^{2}\theta = 3\sin\theta \quad \text{(2)}$$

We need to find all $$\theta \in [0, 2\pi]$$ satisfying both equations and sum them.

First, recall the double-angle identity for cosine: $$\cos 2\theta = 1 - 2\sin^{2}\theta$$.

Substitute this into equation (1):

$$2\sin^{2}\theta = 1 - 2\sin^{2}\theta$$

Bring all terms to one side:

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

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

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

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

Thus, $$\sin\theta = \pm \frac{1}{2}$$.

Now, use equation (2): $$2\cos^{2}\theta = 3\sin\theta$$.

Substitute $$\cos^{2}\theta = 1 - \sin^{2}\theta$$:

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

$$2 - 2\sin^{2}\theta = 3\sin\theta$$

Bring all terms to one side:

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

Multiply both sides by -1:

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

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

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

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

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

So, $$u = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$$ or $$u = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$$.

Since $$|\sin\theta| \leq 1$$, $$\sin\theta = -2$$ is invalid. Thus, only $$\sin\theta = \frac{1}{2}$$.

Now, find all $$\theta \in [0, 2\pi]$$ such that $$\sin\theta = \frac{1}{2}$$.

Sine is positive in the first and second quadrants. The reference angle is $$\frac{\pi}{6}$$.

In first quadrant: $$\theta = \frac{\pi}{6}$$.

In second quadrant: $$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$.

Verify both solutions satisfy the original equations.

For $$\theta = \frac{\pi}{6}$$:

Equation (1): $$2\sin^{2}\left(\frac{\pi}{6}\right) = 2 \left(\frac{1}{2}\right)^2 = 2 \times \frac{1}{4} = \frac{1}{2}$$, $$\cos 2\left(\frac{\pi}{6}\right) = \cos \frac{\pi}{3} = \frac{1}{2}$$, so equal.

Equation (2): $$2\cos^{2}\left(\frac{\pi}{6}\right) = 2 \left(\frac{\sqrt{3}}{2}\right)^2 = 2 \times \frac{3}{4} = \frac{3}{2}$$, $$3\sin\left(\frac{\pi}{6}\right) = 3 \times \frac{1}{2} = \frac{3}{2}$$, so equal.

For $$\theta = \frac{5\pi}{6}$$:

Equation (1): $$2\sin^{2}\left(\frac{5\pi}{6}\right) = 2 \left(\frac{1}{2}\right)^2 = \frac{1}{2}$$, $$\cos 2\left(\frac{5\pi}{6}\right) = \cos \frac{5\pi}{3} = \cos \left(2\pi - \frac{\pi}{3}\right) = \cos \left(-\frac{\pi}{3}\right) = \cos \frac{\pi}{3} = \frac{1}{2}$$, so equal.

Equation (2): $$2\cos^{2}\left(\frac{5\pi}{6}\right) = 2 \left(-\frac{\sqrt{3}}{2}\right)^2 = 2 \times \frac{3}{4} = \frac{3}{2}$$, $$3\sin\left(\frac{5\pi}{6}\right) = 3 \times \frac{1}{2} = \frac{3}{2}$$, so equal.

No other solutions exist in $$[0, 2\pi]$$.

Sum of solutions: $$\frac{\pi}{6} + \frac{5\pi}{6} = \frac{6\pi}{6} = \pi$$.

The sum is $$\pi$$, which corresponds to option C.