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The general value of $\theta $satisfying the equation $2{\sin ^2}\theta - 3\sin \theta - 2 = 0$ is
$n\pi + {( - 1)^n}\frac{\pi }{6}$
$n\pi + {( - 1)^n}\frac{\pi }{2}$
$n\pi + {( - 1)^n}\frac{{5\pi }}{6}$
$n\pi + {( - 1)^n}\frac{{7\pi }}{6}$
Solution
(d) $2{\sin ^2}\theta – 3\sin \theta – 2 = 0$
$ \Rightarrow $ $(2\sin \theta + 1)\,\,(\sin \theta – 2) = 0$
$ \Rightarrow $ $\sin \theta = – \frac{1}{2}$ , $(\therefore \,\sin \theta \ne 2)$
$ \Rightarrow $ $\sin \theta = \sin \left( {\frac{{ – \pi }}{6}} \right)$
$ \Rightarrow $ $\theta = n\pi + {( – 1)^n}\left( {\frac{{ – \pi }}{6}} \right) $
$\Rightarrow \theta = n\pi + {( – 1)^{n + 1}}\frac{\pi }{6}$
$ \Rightarrow $ $\theta = n\pi + {( – 1)^n}\frac{{7\pi }}{6}$, $\left\{ \because {\,\,\frac{{ – \pi }}{6}{\rm{\,\,is \,\,equivalent \,\,to\,\, }}\frac{{7\pi }}{6}} \right\}$.
