Question

The value of $x$ satisfying the inequality $2co{s}^{2}x>\frac{3}{2},\frac{-\pi }{2}\le x\le \frac{3\pi }{2}$ lies in

• A. $(\frac{-\pi }{6},\frac{\pi }{6})\cup (\frac{5\pi }{6},\frac{7\pi }{6})$
• B. $(\frac{\pi }{6},\frac{5\pi }{6})\cup (\frac{7\pi }{6},\frac{\pi }{6})$
• C. $(\frac{-\pi }{2},\frac{-\pi }{6})\cup (\frac{\pi }{6},\frac{7\pi }{6})$
• D. $(\frac{-\pi }{2},\frac{-\pi }{6})\cup (\frac{\pi }{6},\frac{5\pi }{6})$

Answer 1

The correct option is A $(\frac{-\pi }{6},\frac{\pi }{6})\cup (\frac{5\pi }{6},\frac{7\pi }{6})$

$2{\cos }^{2}x>\frac{3}{2}$

$\Rightarrow {\cos }^{2}x>\frac{3}{4}$

$\Rightarrow \cos x>\pm \frac{\sqrt{3}}{2}$

$\cos \theta$ is positive in first and fourth quadrant and negative in second and third quadrant.

$\therefore x\in (-\frac{\pi }{6},\frac{\pi }{6})$ for first and fourth quadrant (when $x>\frac{\sqrt{3}}{2}$)

$\therefore x\in (\pi -\frac{\pi }{6},\pi +\frac{\pi }{6})$ for second and third quadrant. (when $x>-\frac{\sqrt{3}}{2}$)

$\therefore x\in (-\frac{\pi }{6},\frac{\pi }{6})\cup (\frac{5\pi }{6},\frac{7\pi }{6})$