Question

Solution set for the inequality $\frac{5}{4}{\sin }^{2}x+{\sin }^{2}x\cdot {\cos }^{2}x>\cos 2x$ is

• A. $2n\pi +\frac{\pi }{3}<x<2n\pi +\frac{5\pi }{3},n\in Z$
• B. $n\pi +\frac{\pi }{6}<x<n\pi +\frac{5\pi }{6},n\in Z$
• C. $n\pi +\frac{\pi }{6}<x<n\pi +\frac{5\pi }{3},n\in Z$
• D. $n\pi +\frac{\pi }{3}<x<n\pi +\frac{5\pi }{6},n\in Z$

Answer 1

The correct option is B $n\pi +\frac{\pi }{6}<x<n\pi +\frac{5\pi }{6},n\in Z$

$\frac{5}{4}{\sin }^{2}x+{\sin }^{2}x\cdot {\cos }^{2}x>\cos 2x\Rightarrow 5(1-\cos 2x)+2{\sin }^{2}2x>8\cos 2x\Rightarrow 5(1-\cos 2x)+2(1-{\cos }^{2}2x)>8\cos 2x\Rightarrow 2{\cos }^{2}2x+13\cos 2x-7<0\Rightarrow 2{\cos }^{2}2x+14\cos 2x-\cos 2x-7<0\Rightarrow (\cos 2x+7)(2\cos 2x-1)<0\Rightarrow -7<\cos 2x<\frac{1}{2}$
but $-1\le \cos 2x\le 1$
hence $-1\le \cos 2x<\frac{1}{2}$

Now from graph,


$2n\pi +\frac{\pi }{3}<2x<2n\pi +\frac{5\pi }{3}$
$\Rightarrow n\pi +\frac{\pi }{6}<x<n\pi +\frac{5\pi }{6},n\in Z$