Question

Solve : $\cos 2x>\left|\sin x\right|$, $xϵ(-\frac{\pi }{2},\pi )$

• A. $xϵ(-\frac{\pi }{3},\frac{\pi }{3})\cup (\frac{5\pi }{6},\pi )$
• B. $xϵ(-\frac{\pi }{4},\frac{\pi }{4})\cup (\frac{5\pi }{6},\pi )$
• C. $xϵ(-\frac{\pi }{6},\frac{\pi }{6})\cup (\frac{5\pi }{7},\pi )$
• D. $xϵ(-\frac{\pi }{6},\frac{\pi }{6})\cup (\frac{5\pi }{6},\pi )$

Answer 1

Answer: (D)

$xϵ(-\frac{\pi }{6},\frac{\pi }{6})\cup (\frac{5\pi }{6},\pi )$

Draw the graph of $y=\cos 2x$ and $y=\left|\sin x\right|$
Let $\cos 2x=\sin x$.

Then, $2{\sin }^{2}x+\sin x-1=0$ or $\sin x=-1,\frac{1}{2}$
But $\sin x\ne -1$ or $\sin x=\frac{1}{2}$
Clearly, from the graph,

Graphs of $y=\left|\sin x\right|$ and $y=\cos 2x$ intersect at $x=\pm \frac{\pi }{6},\frac{5\pi }{6}$
Thus, the solution set is $xϵ(-\frac{\pi }{6},\frac{\pi }{6})\cup (\frac{5\pi }{6},\pi )$