Question

Complete the set of values of $x$ in $(0,\pi )$ satisfying the inequation $1+{\log }_2\sin x+{\log }_2\sin 3x\ge 0$ is

• A. $[0,\frac{\pi }{6}]$
• B. $[\frac{\pi }{2},\frac{3\pi }{4}]$
• C. $[\frac{\pi }{6},\frac{\pi }{4}]\cup [\frac{3\pi }{4},\frac{5\pi }{6}]$
• D. $[\frac{5\pi }{6},\pi ]$

Answer 1

The correct option is C $[\frac{\pi }{6},\frac{\pi }{4}]\cup [\frac{3\pi }{4},\frac{5\pi }{6}]$

$x\in (0,\pi )\Rightarrow 3x\in (0,3\pi )$
For the inequality to be defined,
$\sin x>0$ and $\sin 3x>0$
$\Rightarrow x\in (0,\frac{\pi }{3})\cup (\frac{2\pi }{3},\pi )\cdots \left(1\right)$

$1+{\log }_2\sin x+{\log }_2\sin 3x\ge 0$
$\Rightarrow {\log }_2\left(2\sin x\sin 3x\right)\ge 0$
$\Rightarrow 2\sin x\sin 3x\ge 1$
$\Rightarrow 2\sin x(3\sin x-4{\sin }^{3}x)\ge 1$
$\Rightarrow 8{\sin }^{4}x-6{\sin }^{2}x+1\le 0$

Let ${\sin }^{2}x=t$
Then $8t^2-6t+1\le 0$
$\Rightarrow (t-\frac{1}{2})(t-\frac{1}{4})\le 0$
$\Rightarrow \frac{1}{4}\le t\le \frac{1}{2}$
$\Rightarrow \frac{1}{4}\le {\sin }^{2}x\le \frac{1}{2}$
$\Rightarrow \frac{1}{2}\le \sin x\le \frac{1}{\sqrt{2}}$
$\Rightarrow x\in [\frac{\pi }{6},\frac{\pi }{4}]\cup [\frac{3\pi }{4},\frac{5\pi }{6}]\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$x\in [\frac{\pi }{6},\frac{\pi }{4}]\cup [\frac{3\pi }{4},\frac{5\pi }{6}]$