Question

The solution set of $x\in (-\pi ,\pi )$ for the inequality $\sin 2x+1\le cosx+2\sin x$ is

• A. $x\in [0,\pi /6]$
• B. $x\in [\frac{\pi }{6},\frac{5\pi }{6}]\sim \left\{0\right\}$
• C. $x\in [\frac{\pi }{6},\frac{5\pi }{6}]$
• D. $noneofthese$

Answer 1

The correct option is A $x\in [0,\pi /6]$

$\sin 2x+1\le cosx+2sinx$

$\Rightarrow 2sinxcosx+1-cosx-2sinx\le 0$

$\Rightarrow 2sinx(cosx-1)-1(cosx-1)\le 0$

$\Rightarrow (2sinx-1)(cosx-1)\le 0$

$x\in [0,\pi /6]$