Question

The values of $x$ in the interval $[0,\pi ]$ such that $\sin 2x=\frac{\sqrt{3}}{2}$ are

• A. $\frac{\pi }{6},\frac{\pi }{3}$
• B. $\frac{\pi }{6},\frac{2\pi }{3}$
• C. $\frac{\pi }{3},\frac{2\pi }{3}$
• D. $\frac{\pi }{6},\frac{5\pi }{6}$
• E. $\frac{\pi }{3},\frac{5\pi }{6}$

Answer 1

Answer: (A)

$\sin 2x=\frac{\sqrt{3}}{2}\forall x\in [0,\pi ]$
$\text{ If }x\in [0,\pi ],2x\in [0,2\pi ]$
$\sin 2x=\sin \frac{\pi }{3}\text{ or }\sin 2x=\sin \left(\pi -\frac{\pi }{3}\right)$
$\Rightarrow 2x=\frac{\pi }{3},\frac{2\pi }{3}$
$x=\frac{\pi }{6},\frac{\pi }{3}$