Question

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

Answer 1

Given that

$r\sin \theta =3$ ....$\left(1\right)$

$r=4(1+\sin \theta )$ ....$\left(2\right)$

From $\left(1\right)$ we have $\sin \theta =\frac{3}{r}$

Substitute this in $\left(2\right)$, we get

$\Rightarrow r=4+4\times \left(\frac{3}{r}\right)$

$\Rightarrow r=4+\frac{12}{r}$

$\Rightarrow r^2=4r+12$

$\Rightarrow r^2-4r-12=0$

$\Rightarrow (r-6)(r+2)=0$

$\Rightarrow r=6,r=-2$

$\Rightarrow r=6\Rightarrow \sin \theta =\frac{3}{6}=\frac{1}{2}$

So, $\theta =\frac{\pi }{6},\frac{5\pi }{6}$

$\Rightarrow r=-2\Rightarrow \sin \theta =\frac{-3}{2}$ which is not possible.

Therefore, $\theta =\frac{\pi }{6},\frac{5\pi }{6}$