Question

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

Answer 1

Eliminating r between the given equations, we get $\frac{\sqrt{3}}{\sin \theta }+4\sin \theta =2(\sqrt{3}+1)$
or $4{\sin }^2\theta -2\sqrt{3}\sin \theta -2\sin \theta +\sqrt{3}=0$
$2\sin \theta (2\sin \theta -\sqrt{3})-1.(2\sin \theta -\sqrt{3})=0$
$\therefore (2\sin \theta -1)(2\sin \theta -\sqrt{3})=0$
$\sin \theta =\frac{1}{2}=\sin \frac{\pi }{6}$ or $\sin (\pi -\frac{\pi }{6})$
$\sin \theta =\frac{\sqrt{3}}{2}=\sin \frac{\pi }{3}$ or $\sin (\pi -\frac{\pi }{3})$
$\therefore \theta =\pi /6,\pi /3,5\pi /6,2\pi /3,0\le \theta \le 2\pi$.