# How do you find the points of intersection of r=1/sin(Theta) and r=3-2sin(Theta) ?

## $r = 3 - 2 \sin \left(\Theta\right)$

May 17, 2018

Points of intersection in the first quadrant on the polar co-ordinates are:
$\left(2 , \frac{\pi}{6}\right)$ and $\left(1 , \frac{\pi}{2}\right)$

#### Explanation:

The simplest way of finding the point of intersection is to substitute the value of $\theta$ from the first equation into the second equation. Here, as $\theta$ is in terms of its sine, we will substitute the value of $\sin \theta$ from the first equation into the second one.

we know,
$r = \frac{1}{\sin} \theta$
$: , \sin \theta = \frac{1}{r}$ . . . . . . . . [i]

Equation 2:
$r = 3 - 2 \sin \theta$ . . . . . . . . . [ii]

Now, at point of intersection, the 2 equations will have the same value of r and $\theta$

$\therefore$ Substituting value of $\sin \theta$ from eq.[i] into [ii],

$r = 3 - 2 \left(\frac{1}{r}\right)$

$\therefore {r}^{2} = 3 r - 2$ . . . . . . .[ Multiplying both sides by r]
$\therefore {r}^{2} - 3 r + 2 = 0$

Now, this is a quadratic equation, which can be factorized as:
${r}^{2} - 2 r - r + 2 = 0$

$\therefore r \left(r - 2\right) - \left(r - 2\right) = 0$

$\therefore \left(r - 1\right) \left(r - 2\right) = 0$

which means,
$r - 1 = 0$
or
$r - 2 = 0$

$\therefore r = 1$ or
$r = 2$

if r=1,
from eq. [i],

$\sin \theta = \frac{1}{1}$ i.e. $\theta = \frac{\pi}{2}$

if r=2,
from eq. [i],

$\sin \theta = \frac{1}{2}$ i.e. $\theta = \frac{\pi}{6}$

Thus, in the first quadrant, the given curves intersect at 2 points,
$\left(2 , \frac{\pi}{6}\right)$ and $\left(1 , \frac{\pi}{2}\right)$