Question

If $|1-\frac{|\sin x|}{1+|\sin x|}|\ge \frac{2}{3}$, then $x$ lies in

• A. $[-\frac{\pi }{6},\frac{\pi }{6}]$
• B. $[\frac{5\pi }{6},\frac{7\pi }{6}]$
• C. $[-\frac{7\pi }{6},-\frac{5\pi }{6}]$
  
• D. $[\frac{\pi }{6},\frac{5\pi }{6}]$
  

Answer 1

Answer: (A), (B), (C)

The correct options are
A $[-\frac{\pi }{6},\frac{\pi }{6}]$
B $[\frac{5\pi }{6},\frac{7\pi }{6}]$
C $[-\frac{7\pi }{6},-\frac{5\pi }{6}]$


$|1-\frac{|\sin x|}{1+|\sin x|}|\ge \frac{2}{3}\Rightarrow \frac{1}{|1+|\sin x||}\ge \frac{2}{3}$
$\because 1+|\sin x|\ge 1\forall x\in R$
$\therefore \frac{1}{1+|\sin x|}\ge \frac{2}{3}\Rightarrow |\sin x|\le \frac{1}{2}$


From the graph,
$|\sin x|\le \frac{1}{2}\Rightarrow x\in [-\frac{\pi }{6},\frac{\pi }{6}]$
$\because$ period of $y=|\sin x|$ is $\pi$
$\therefore x\in \bigcup _{n\in Z}[n\pi -\frac{\pi }{6},n\pi +\frac{\pi }{6}]$