If A = $\{\theta : 2 cos^2 \theta + sin \theta ≤ 2\}$  and  B = $\{\theta : \frac{\pi}{2} ≤ \theta ≤ \frac{3\pi}{2}$, then A ∩ B is equal to

$\left\{\theta: \frac{\pi}{2} \leq \theta \leq \frac{5 \pi}{6}\right.$  or  $\left.\leq \theta \leq \frac{3 \pi}{2}\right\}$

Let : $2 cos^2 \theta + sin \theta ≤ 2$  and  $\frac{3\pi}{2}$

$\Rightarrow 2-2 \sin ^2 \quad \theta+\sin \theta \leq 2$

$\Rightarrow 2 \sin ^2 \quad \theta-\sin \theta \geq 0 \Rightarrow \sin \theta(2 \sin \theta-1) \geq 0$

Case I. $\sin \theta \geq 0,2 \sin \theta-1 \geq 0$

∴ $\sin \theta \geq 0, \sin \theta \geq \frac{1}{2} \Rightarrow \sin \theta \geq \frac{1}{2}$

$\Rightarrow \frac{\pi}{2} \leq \theta \leq \frac{5 \pi}{6}$

Case II. $\sin \theta \leq 0,2 \sin \theta-1 \leq 0$

$\sin \theta \leq 0, \sin \theta \leq \frac{1}{2} \Rightarrow \sin \theta \leq 0$

$\Rightarrow \pi \leq \theta \leq \frac{3 \pi}{2}$

∴  $A \cap B=\left\{\theta: \frac{\pi}{2} \leq \theta \leq \frac{5 \pi}{6} \text { or } \pi \leq \theta \leq \frac{3 \pi}{2}\right\}$

Hence (3) is the correct answer.