Number of Solutions of Simultaneous Trigonometric Equations

To find the number of solutions of the pair of simultaneous trigonometric equations in the interval $[0, 2\pi]$:

First Equation:
$2\sin^2\theta - \cos 2\theta = 0$

Using the double-angle identity $\cos 2\theta = 1 - 2\sin^2\theta$, we substitute this into the equation:
$2\sin^2\theta - (1 - 2\sin^2\theta) = 0$
$4\sin^2\theta - 1 = 0$
$\sin^2\theta = \frac{1}{4} \implies \sin\theta = \pm\frac{1}{2}$

Second Equation:
$2\cos^2\theta - 3\sin\theta = 0$

Using the identity $\cos^2\theta = 1 - \sin^2\theta$, we rewrite the equation in terms of $\sin\theta$:
$2(1 - \sin^2\theta) - 3\sin\theta = 0$
$2 - 2\sin^2\theta - 3\sin\theta = 0$
$2\sin^2\theta + 3\sin\theta - 2 = 0$

Factoring the quadratic equation:
$(2\sin\theta - 1)(\sin\theta + 2) = 0$

Since $-1 \le \sin\theta \le 1$, the factor $\sin\theta + 2 = 0$ yields no real solution. Thus:
$\sin\theta = \frac{1}{2}$

Simultaneous Solutions:
For both equations to be satisfied simultaneously, $\theta$ must satisfy:
$\sin\theta = \frac{1}{2}$

In the interval $[0, 2\pi]$, the solutions for $\sin\theta = \frac{1}{2}$ are:
$\theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{5\pi}{6}$

Thus, there are exactly two solutions.

Correct Option: C