We need the common solutions of $2\sin^2\theta=\cos2\theta \quad \text{and} \quad 2\cos^2\theta=3\sin\theta,$ for $\theta\in[0,2\pi]$.

Solve the first equation.

Using $\cos2\theta=1-2\sin^2\theta,$ we get $2\sin^2\theta=1-2\sin^2\theta$ $4\sin^2\theta=1$ $\sin^2\theta=\frac14$ $\sin\theta=\pm\frac12.$

Solve the second equation.

Using $\cos^2\theta=1-\sin^2\theta,$ $2(1-\sin^2\theta)=3\sin\theta$ $2-2\sin^2\theta=3\sin\theta$ $2\sin^2\theta+3\sin\theta-2=0.$

Let $x=\sin\theta$. Then $2x^2+3x-2=0$ $(2x-1)(x+2)=0.$

So, $x=\frac12 \quad \text{or} \quad x=-2.$

Since $\sin\theta\in[-1,1]$, $x=-2$ is impossible. Hence $\sin\theta=\frac12.$

For both equations to hold simultaneously, we must have $\sin\theta=\frac12.$

In $[0,2\pi]$, this gives $\theta=\frac{\pi}{6},\ \frac{5\pi}{6}.$

Sum of all such values: $\frac{\pi}{6}+\frac{5\pi}{6}=\pi.$

Therefore, the correct option is $\boxed{\text{A }(\pi)}.$