Question

If 4${sin}^{2}x-8sinx+3\le 0,0\le x\le 2\pi ,xϵ$

• 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 $[\frac{\pi }{6},\frac{5\pi }{6}]$

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

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

$2\sin x(2\sin x-3)-1(2\sin x-3)\le 0$

$\Rightarrow 2\sin x-3=0,2\sin x-1=0$

$\Rightarrow \sin x=\frac{3}{2},\sin x=\frac{1}{2}$

$\sin x\ne \frac{3}{2}$ since it is not in range of $\sin x$

$\therefore \sin x=\frac{1}{2}$

$\therefore x=\frac{\pi }{6}$

$\sin x=\sin \frac{\pi }{6}=\frac{1}{2}$

$\sin x=\sin (\pi -\frac{\pi }{6})=\frac{1}{2}$

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