Question

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

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

Answer 1

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

$1+{\log }_2\sin x+{\log }_2\sin 3x\ge 0$
For $\log$ function to be defined, we get
$\sin x>0,\sin 3x>0\sin x>0\Rightarrow x\in (0,\pi )\cdots \left(1\right)\sin 3x>0\Rightarrow 3x\in (0,\pi )\cup (2\pi ,3\pi )\Rightarrow x\in (0,\frac{\pi }{3})\cup (\frac{2\pi }{3},\pi )\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we get
$x\in (0,\frac{\pi }{3})\cup (\frac{2\pi }{3},\pi )$

Now,
${\log }_2\left(2\sin x\sin 3x\right)\ge 0$
$\Rightarrow 2\sin x\sin 3x\ge 1\Rightarrow 2{\sin }^{2}x(3-4{\sin }^{2}x)\ge 1\Rightarrow 8{\sin }^{4}x-6{\sin }^{2}x+1\le 0\Rightarrow (2{\sin }^{2}x-1)(4{\sin }^{2}x-1)\le 0\Rightarrow \frac{1}{4}\le {\sin }^{2}x\le \frac{1}{2}\Rightarrow \frac{1}{2}\le \sin x\le \frac{1}{\sqrt{2}}(\because \sin x>0)\therefore x\in [\frac{\pi }{6},\frac{\pi }{4}]\cup [\frac{3\pi }{4},\frac{5\pi }{6}]$