Question

Solve $4{\sin }^{2}x-8\sin x+3\ge 0,xϵ[0,2\pi ]$

Answer 1

$4{\sin }^{2}x-8\sin x+3\ge 0,x\in [0,2\pi ]$

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

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

$\underset{alwaysnegative}{(2\sin x-1)(2\sin x-3)\ge 0}$

$2\sin x-1\le 0$

$sinx\le \frac{1}{2}$

$x\in [0,\frac{\pi }{6}]v[\frac{5\pi }{6},\pi ]$