Question

Solve $r\sin \theta =3,r=4(1+\sin \theta ),0\le \theta \le 2\pi$.

Answer 1

Eliminating r, we get $4{\sin }^2\theta +4\sin \theta -3=0$
or $(2\sin \theta +3)(2\sin \theta -1)=0$
$\therefore \sin \theta =\frac{1}{2}=\sin \frac{\pi }{6}$
$\theta =\pi /6,\pi -\pi /6$ for $0\le \theta \le 2\pi$.
The other value is rejected as $\sin \theta$ cannot be greater than $1$.