Question

If $3 \tan ^{2} \theta-2 \sin \theta=0$, then $\theta=$

• A. $2 n \pi$
• B. $n \pi+(-1)^{n} \frac{\pi}{6}$
• C. $n \pi-(-1)^{n} \frac{\pi}{6}$
• D. $n \pi+\frac{\pi}{3}$

Answer 1

Answer: (B)

$3 \frac{\sin ^{2} \theta}{\cos ^{2} \theta}-2 \sin \theta=0, \cos \theta \neq 0$
$\Rightarrow 3 \sin \theta^{2}-2 \sin \theta\left(1-\sin \theta^{2}\right)=0$
$\Rightarrow 3 \sin \theta^{2}-2 \sin \theta\left(1-\sin \theta^{2}\right)=0$
$\Rightarrow \sin \theta\left(2 \sin \theta^{2}+3 \sin \theta-2\right)=0$
$\Rightarrow \sin \theta(2 \sin \theta-1)(\sin \theta+2)=0$
$\Rightarrow \sin \theta=0,1-2($ rejected $)$
$\Rightarrow \theta=n \pi, n \pi+(-1)^{n} \frac{\pi}{6}$