Question

If $|2\sin \theta -\text{cosec}\theta |\ge 1$, then $\theta \in$

• A. $\bigcup _{n\in Z}[n\pi -\frac{\pi }{6},n\pi +\frac{\pi }{6}]\cup \left\{\frac{(2n+1)\pi }{2}\right\}-\left\{n\pi \right\}$
• B. $\bigcup _{n\in Z}[n\pi -\frac{\pi }{3},n\pi +\frac{\pi }{3}]\cup \left\{\frac{(2n+1)\pi }{2}\right\}-\left\{n\pi \right\}$
• C. $\bigcup _{n\in Z}(n\pi -\frac{\pi }{3},n\pi +\frac{\pi }{3})$
• D. $\bigcup _{n\in Z}(n\pi -\frac{\pi }{6},n\pi +\frac{\pi }{6})$

Answer 1

The correct option is A $\bigcup _{n\in Z}[n\pi -\frac{\pi }{6},n\pi +\frac{\pi }{6}]\cup \left\{\frac{(2n+1)\pi }{2}\right\}-\left\{n\pi \right\}$

$\text{cosec}\theta$ is not defined for $n\pi ,\forall n\in Z$
$|2\sin \theta -\text{cosec}\theta |\ge 1$
$\Rightarrow |2{\sin }^2\theta -1|\ge |\sin \theta |$
$\Rightarrow 2{\cos }^{2}2\theta \ge 1-\cos 2\theta$
$\Rightarrow 2{\cos }^{2}2\theta +\cos 2\theta -1\ge 0$
$\Rightarrow (2\cos 2\theta -1)(\cos 2\theta +1)\ge 0$
$\Rightarrow \cos 2\theta \ge \frac{1}{2}\text{or}\cos 2\theta \le -1$

Now, using the graph $2\theta \in [-\frac{\pi }{3},\frac{\pi }{3}]$ and period of $\cos 2\theta$ is $\pi .$
If $\cos 2\theta \le -1,$ then $\theta =\frac{(2n+1)\pi }{2}$
Hence, $\theta \in \bigcup _{n\in Z}[n\pi -\frac{\pi }{6},n\pi +\frac{\pi }{6}]\cup \left\{\frac{(2n+1)\pi }{2}\right\}-\left\{n\pi \right\}$