Question

Find the values of $\theta$ which satisfy $r\sin \theta =3$ and $r=4(1+\sin \theta )$, $0\le \theta \le 2\pi$

• A. $\theta =\frac{\pi }{6}$
• B. $\theta =\frac{5\pi }{6}$
• C. $\theta =\frac{\pi }{3}$
• D. $\theta =\frac{2\pi }{3}$

Answer 1

Answer: (A), (B)

The correct options are
A $\theta =\frac{5\pi }{6}$
B $\theta =\frac{\pi }{6}$

We have, $0\le \theta \le 2\pi$

Eliminating r, we have $4{\sin }^2\theta +4\sin \theta -3=0$

$\Rightarrow$ $4{\sin }^2\theta -2\sin \theta +6\sin \theta -3=0$

$\Rightarrow$ $2\sin \theta (2\sin \theta -1)+3(2\sin \theta -1)=0$

$\Rightarrow$ $\sin \theta =\frac{1}{2},-\frac{3}{2}$(not possible)

$\Rightarrow$ $\theta =\frac{\pi }{6},\pi -\frac{\pi }{6}=\frac{5\pi }{6}$ in the given interval.