Question

If $|2\sin \theta -\text{cosec}\theta |\ge 1$, then complete set of values of $\theta$ is
(where $n\in Z$)

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

Answer 1

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

Given : $|2\sin \theta -\text{cosec}\theta |\ge 1$
For $\text{cosec}\theta$ to be defined, we get
$\theta \ne n\pi ,\forall n\in Z$
Now,
$|2\sin \theta -\text{cosec}\theta |\ge 1\Rightarrow |2{\sin }^2\theta -1|\ge |\sin \theta |\Rightarrow |-\cos 2\theta |\ge |\sin \theta |$
Squaring on both sides, we get
$\Rightarrow {\cos }^{2}2\theta \ge {\sin }^2\theta \Rightarrow 2{\cos }^{2}2\theta \ge 1-\cos 2\theta \Rightarrow 2{\cos }^{2}2\theta +\cos 2\theta -1\ge 0\Rightarrow (2\cos 2\theta -1)(\cos 2\theta +1)\ge 0\Rightarrow \cos 2\theta \ge \frac{1}{2}\text{or}\cos 2\theta \le -1$
As $\cos 2\theta \in [-1,1]$, so
$\cos 2\theta \ge \frac{1}{2},\cos 2\theta =-1$


For $\cos 2\theta \ge \frac{1}{2}$, using the graph of $\cos x$, we get
$2\theta \in [-\frac{\pi }{3},\frac{\pi }{3}]\Rightarrow 2\theta \in [2n\pi -\frac{\pi }{3},2n\pi +\frac{\pi }{3}]\Rightarrow \theta \in [n\pi -\frac{\pi }{6},n\pi +\frac{\pi }{6}]$
Also,
$\cos 2\theta =-1\Rightarrow 2\theta =(2n+1)\pi \Rightarrow \theta =\left\{\frac{(2n+1)\pi }{2}\right\}$

Hence, $\theta \in [n\pi -\frac{\pi }{6},n\pi +\frac{\pi }{6}]\cup \left\{\frac{(2n+1)\pi }{2}\right\}-\left\{n\pi \right\},n\in Z$