Question

The set of values of $x$ in $(0,\pi )$ satisfying the equation $1+{\log }_2\sin x+{\log }_2\sin 3x\ge 0$ can't lie in is

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

Answer 1

Answer: (A), (B), (D)

$(\frac{2\pi }{3},\frac{3\pi }{4})$
$(\frac{\pi }{2},\frac{2\pi }{3})$
$(\frac{\pi }{3},\frac{2\pi }{3}]$

$1+{\log }_2\sin x+{\log }_2\sin 3x\ge 0={\log }_2\left(2\sin x\sin 3x\right)\ge 0$
$\Rightarrow 2\sin x\sin 3x\ge 1$
$\Rightarrow 2{\sin }^{2}x(3-4{\sin }^{2}x)\ge 1$
Let ${\sin }^{2}x$ be $t$
Thus, $6t-8t^2-1\ge 0\Rightarrow 8t^2-6t+1\le 0$
$\Rightarrow (2t-1)(4t-1)\le 0$ or $t\in [\frac{1}{4},\frac{1}{2}]$
So, $\sin x\in [\frac{1}{2},\frac{1}{\sqrt{2}}]$
So, the range $(0,\pi ),x\in [\frac{\pi }{6},\frac{\pi }{4}]\cup [\frac{3\pi }{4},\frac{5\pi }{6}]$