Question

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

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