Question

The general solution of $3{\tan }^2\theta -2\sin \theta =0$ is

• A. $\theta =n\pi :n\in I$
• B. $\theta =n\pi +(-1{)}^n\cdot \frac{\pi }{6};n\in I$
• C. $0=n\pi +(-1{)}^n\cdot \frac{\pi }{3};n\in I$
• D. Both $\left(1\right)$ and $\left(2\right)$

Answer 1

The correct option is D Both $\left(1\right)$ and $\left(2\right)$

Given $3{\tan }^2\theta -2\sin \theta =0$

$⟹3\frac{{\sin }^2\theta }{{\cos }^2\theta }-2\sin \theta =0$

$⟹\sin \theta (\frac{3\sin \theta }{{\cos }^2\theta }-2)=0$

$⟹\frac{3\sin \theta }{{\cos }^2\theta }-2=0$ or $\sin \theta =0$

$⟹3\sin \theta =2(1-{\sin }^2\theta )$ or $\theta =n\pi$

$⟹2{\sin }^2\theta +3\sin \theta -2=0$

$⟹2{\sin }^2\theta +4\sin \theta -\sin \theta -2=0$

$⟹(2\sin \theta -1)(\sin \theta +2)=0$

$⟹\sin \theta =\frac{1}{2}⟹\theta =n\pi +(-1{)}^n\frac{\pi }{6}$

So $\theta =n\pi +(-1{)}^n\frac{\pi }{6}\cup n\pi ;n\in I$