Question

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

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

Answer 1

The correct option is D $[\frac{\pi }{6},\frac{5\pi }{6}]$

$4{\sin }^{2}x-8\sin x+3\le 0$

$\Rightarrow (2\sin x-3)(2\sin x-1)\ge 0$

But we know, $-1\le \sin x\le 1\Rightarrow 2\sin x-3<0$

$\Rightarrow 2\sin x-1\ge 0\Rightarrow \frac{1}{2}\le \sin x\le 1$

Hence solution in given interval is,

$x\in [\frac{\pi }{6},\frac{5\pi }{6}]$