Question

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

(A) $n \pi+(-1)^{n+1} \frac{\pi}{6}$

(B) $n \pi+(-1)^n \frac{\pi}{2}$

(C) $n \pi+(-1)^n \frac{5 \pi}{6}$

(D) $n \pi+(-1)^n \frac{7 \pi}{6}$

Answer 1

SOLUTION — Given equation can be rewritten as

$\begin{array}{lc} & (2 \sin \theta+1)(\sin \theta-2)=0 \\\Rightarrow & \sin \theta=-\frac{1}{2} \\\Rightarrow & \sin \theta=\sin \left(-\frac{\pi}{6}\right) \\\Rightarrow & \theta=n \pi+(-1)^n\left(-\frac{\pi}{6}\right) \\\Rightarrow & \theta=n \pi+(-1)^{n+1} \frac{\pi}{6}\end{array}$

So, The correct option will be (A).