Question

The general value of $\theta$ satisfying equation
$2{\sin }^2\theta -3\sin \theta -2=0$ is

• A. $n\pi +{(-1)}^n\frac{\pi }{6}$
• B. $n\pi +{(-1)}^n\frac{\pi }{2}$
• C. $n\pi +{(-1)}^n\frac{5\pi }{6}$
• D. $n\pi +{(-1)}^{n+1}\frac{\pi }{6}$

Answer 1

The correct option is D $n\pi +{(-1)}^{n+1}\frac{\pi }{6}$

we have, $2{\sin }^2\theta -3\sin \theta -2=0$
$\Rightarrow (2\sin \theta +1)(\sin \theta -2)=0$
$\Rightarrow \sin \theta =-\frac{1}{2}[\because \sin \theta \ne 2]$
$\Rightarrow \sin \theta =\sin (-\frac{\pi }{6})$
$\Rightarrow \theta =n\pi +{(-1)}^n(-\frac{\pi }{6})$
$\Rightarrow \theta =n\pi +{(-1)}^{n+1}\left(\frac{\pi }{6}\right)$