Question

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

Answer 1

$3\cdot \frac{{\sin }^2\theta }{{\cos }^2\theta }-2\sin \theta =0,\cos \theta \ne 0$.
Multiplying throughout by ${\cos }^2\theta$ and then changing it into ${\sin }^2\theta$. we get
$3{\sin }^2\theta -2\sin \theta (1-{\sin }^2\theta )=0$
$\sin \theta (3\sin \theta -2+2{\sin }^2\theta )=0$
$\sin \theta (2{\sin }^2\theta +4\sin \theta -\sin \theta -2)=0$
$\sin \theta (\sin \theta +2)(2\sin \theta -1)=0$
$\therefore \sin \theta =0$, $\therefore \theta =n\pi$
$\sin \theta =1/2=\sin \left(\pi /6\right)$,
$\therefore \theta =n\pi +(-1{)}^n\pi /6$
$\sin \theta =-2$(rejected as $\sin \theta$ cannot be greater than one numerically).