Question

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

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

Answer 1

The correct option is D

$x\in [\frac{\pi }{6},\frac{\pi }{4}]\cup [\frac{3\pi }{4},\frac{5\pi }{6}]$

$1+{\log }_2\sin x+{\log }_2\sin 3x\ge 0$

$\therefore \sin x>0,x\in (0,\pi )$ and

$\sin 3x>0,3x\in (0,\pi )\cup (2\pi ,3\pi )\Rightarrow x\in (0,\frac{\pi }{3})\cup (\frac{2\pi }{3},\pi )$ ... (1)

$1+{\log }_2\sin x+{\log }_2\sin 3x\ge 0$

$\Rightarrow {\log }_{2}2+{\log }_2\sin x+{\log }_2\sin 3x\ge 0\Rightarrow {\log }_2\left(2\sin x\sin 3x\right)\ge 0$

$\Rightarrow 2\sin x\sin 3x\ge 1$

$\Rightarrow 2\sin x(3\sin x-4{\sin }^{3}x)\ge 1$

$\Rightarrow 6{\sin }^{2}x-8{\sin }^{4}x\ge 1$

$\Rightarrow 6{\sin }^{2}x-8{\sin }^{4}x-1\ge 0\Rightarrow 8{\sin }^{4}x-6{\sin }^{2}x+1\le 0$

$\Rightarrow (2{\sin }^{2}x-1)(4{\sin }^{2}x-1)\le 0$

$\Rightarrow (\sin x-\frac{1}{\sqrt{2}})(\sin x+\frac{1}{\sqrt{2}})(\sin x-\frac{1}{2})(\sin x+\frac{1}{2})\le 0$

Using wavy curver method we get that $\frac{1}{2}\le \sin x\le \frac{1}{\sqrt{2}}$ and

$\frac{-1}{\sqrt{2}}\le \sin x\le \frac{-1}{2}$.

But $\sin x>0$.

Hence $\frac{1}{2}\le \sin x\le \frac{1}{\sqrt{2}}$

$\therefore x\in (\frac{\pi }{6},\frac{\pi }{4})\cup (\frac{3\pi }{4},\frac{5\pi }{6})$ ... (2)

From (1) and (2)

$x\in (\frac{\pi }{6},\frac{\pi }{4})\cup (\frac{3\pi }{4},\frac{5\pi }{6})$