Question

The solution set of $x\in \left(- \pi , \pi \right)$ for the inequality $sin 2 x + 1\leq cos ⁡ x+2sin ⁡ x$ is

• A. $x\in \left[0 , \frac{\pi }{6}\right]$
• B. $x\in \left[\frac{\pi }{6} , \frac{5 \pi }{6}\right]\cup\left\{0\right\}$
• C. $x\in \left[- \frac{\pi }{6} , \frac{5 \pi }{6}\right]$
• D. $x\in \left[- \frac{\pi }{2} , \frac{\pi }{2}\right]$

Answer 1

Answer: (B)

$2sin xcos ⁡ x+1-cos ⁡ x-2sin ⁡ x\leq 0$
$\left(2 sin x - 1\right)\left(cos ⁡ x - 1\right)\leq 0$
Case I: $cos x=1\Rightarrow x=0$
Case II: otherwise $cos x < 1$
$\Rightarrow 2sin x-1\geq 0\Rightarrow sin ⁡ x\geq \frac{1}{2}$
$\Rightarrow x\in \left[\frac{\pi }{6} , \frac{5 \pi }{6}\right]$
Hence, from case I and case II,
$x\in \left[\frac{\pi }{6} , \frac{5 \pi }{6}\right]\cup\left\{0\right\}$