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. $[\frac{\pi }{6},2\pi ]$
• D. $[\frac{\pi }{6},\frac{5\pi }{6}]$

Answer 1

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

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

$\Rightarrow 4{\sin }^{2}x-2\sin x-6\sin x+3⩽0$

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

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

$\sin x\in [\frac{1}{2},\frac{3}{2}]$

but me know $\sin x\in [-1,1]$

so Required values of $\sin x$ lies in $[\frac{1}{2},1]$

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

for

$0⩽x⩽2\pi$

Answer: Option (D)