Question

If $4{\sin }^{2}x-8\sin x+3=0,$ $0\le x\le 2\pi ,$ then the solution set for $x$ is

• A. $\left[0,\frac{\pi }{6}\right]$
• B. $\left[0,\frac{5\pi }{6}\right]$
• C. $\left[\frac{5\pi }{6},2\pi \right]$
• D. $\left[\frac{5\pi }{6},\frac{\pi }{6}\right]$

Answer 1

Answer: (D)

We have,
$4{\sin }^{2}x-8\sin x+3\le 0,0\le x\le 2\pi$
$\Rightarrow$ $(2\sin x-1)(2\sin x-3)\le 0$
But $2\sin x-3$ is always negative, as $\sin x\le 1$
$\therefore$ $2\sin x-1\ge 0$
$\Rightarrow$ $\sin x\ge \frac{1}{2}$
$\therefore$ $\frac{\pi }{6}\le x\le \frac{5\pi }{6}$
Hence, $x\in \left[\frac{\pi }{6},\frac{5\pi }{6}\right]$