Question

The number of solution of the inequality $2{\cos }^{2}x+\sin x\le 2$ in $x\in [0,\frac{\pi }{6}]$ is

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

The correct option is C $2$

$2{\cos }^{2}x+\sin x\le 2\Rightarrow 2(1-{\sin }^{2}x)+\sin x\le 2\Rightarrow 2{\sin }^{2}x-\sin x\ge 0\Rightarrow \sin x(2\sin x-1)\ge 0\Rightarrow \sin x\le 0,\sin x\ge \frac{1}{2}$
in $[0,\frac{\pi }{6}],\sin x\in [0,\frac{1}{2}]$
$\therefore \sin x=0\text{or}\sin x=\frac{1}{2}\Rightarrow x=0\text{or}\frac{\pi }{6}$