The number of solutions of the pair of equations $2\sin^{2}\theta - \cos 2\theta = 0$, $2\cos^{2}\theta - 3\sin\theta = 0$ in the interval $[0, 2\pi]$ is

$2\sin^{2}\theta - \cos 2\theta = 0$ $2\cos^{2}\theta - 3\sin\theta = 0$ $\text{To find: number of common solutions in } [0, 2\pi]$ $\cos 2\theta = 1 - 2\sin^{2}\theta$ $2\sin^{2}\theta - (1 - 2\sin^{2}\theta) = 0$ $4\sin^{2}\theta - 1 = 0 \ \Rightarrow \ \sin^{2}\theta = \dfrac14$ $\sin\theta = \pm\dfrac12$ $\cos^{2}\theta = 1 - \sin^{2}\theta$ $2(1 - \sin^{2}\theta) - 3\sin\theta = 0$ $2\sin^{2}\theta + 3\sin\theta - 2 = 0$ $(2\sin\theta - 1)(\sin\theta + 2) = 0$ $\sin\theta = \dfrac12 \quad (\sin\theta = -2 \text{ rejected})$ $1.\ \text{solve each equation}$ $2.\ \text{take the common solutions}$ $3.\ \text{count them in } [0, 2\pi]$ $\text{Eq.1: } \sin\theta = \pm\dfrac12$ $\text{Eq.2: } \sin\theta = \dfrac12$ $\text{Common: } \sin\theta = \dfrac12$ $\sin\theta = \dfrac12 \ \Rightarrow \ \theta = \dfrac{\pi}{6}, \ \dfrac{5\pi}{6} \ \in [0, 2\pi]$ $\text{At } \theta = \dfrac{\pi}{6}: \ \sin\theta = \dfrac12, \ \cos 2\theta = \cos\dfrac{\pi}{3} = \dfrac12$ $\text{Eq.1: } 2\!\left(\dfrac14\right) - \dfrac12 = 0 \ \checkmark$ $\text{Eq.2: } 2\!\left(\dfrac34\right) - 3\!\left(\dfrac12\right) = 0 \ \checkmark$ $\text{Number of common solutions} = 2$ $\boxed{\text{Two solutions} \ \Rightarrow \ \text{option (3)}}$