Question

The values of x in the interval $[0,\pi]$ such that $sin 2x = \frac{\sqrt3}2$

• 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

The Correct Option is A

Given equation:

\(\sin 2x = \frac{\sqrt{3}}{2}\) 

We know that:

\(\sin \theta = \frac{\sqrt{3}}{2}\) at \(\theta = \frac{\pi}{3}, \frac{2\pi}{3} \) in the interval \([0, \pi]\).

Setting \(2x\) equal to these values:

\(2x = \frac{\pi}{3}, \frac{2\pi}{3}\)

Solving for \(x\):

\(x = \frac{\pi}{6}, \frac{\pi}{3}\)

Thus, the correct answer is:

\(\frac{\pi}{6}, \frac{\pi}{3}\)