If $A=\left\{x: \cos x>-\frac{1}{2}\right.$ and $\left.0 \leq x \leq \pi\right\}, B=\left\{x: \sin x>\frac{1}{2}\right.$ and $\left.\frac{\pi}{3} \leq x \leq \pi\right\}$ Then $A \cap B=$

$\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right)$

$A=\{x: \cos x>-1 / 2$  and  $0 \leq x \leq \pi\}$

$B=\{x: \sin x>1 / 2$  and  $\pi / 3 \leq x \leq \pi\}$

$\cos x>-\frac{1}{2}$

$\Rightarrow x<\frac{2 \pi}{3}$

$A=\left[0, \frac{2 \pi}{3}\right)$

$\sin x>\frac{1}{2} \Rightarrow x>\frac{\pi}{6}, x<\frac{5 \pi}{6}$

$B=\left(\frac{\pi}{6}, \frac{5 \pi}{3}\right)$

$A \cap B=\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right)$

Hence (3) is the correct answer.