I need to convert the polar equation $r = 3\sin(\theta)$ to cartesian form. Using the equations $x^2+y^2=r^2$, $x = r\cos(x)$ and $y = r\sin(x)$,
$$ r = 3\sin(\theta)\\ \implies r^2 = 3r\sin(\theta)\\ \iff x^2 + y^2 = 3x\\ \iff (x-\frac 32)^2 +y^2 = (\frac 32)^2 $$ I understand that all $(r,\theta)$ that satisfy $r = 3\sin(\theta)$ also satisfy $r^2 = 3r\sin(\theta)$, so $r = 3\sin(\theta) \implies r^2 = 3r\sin(\theta)$.
The solutions manual says that $r = 3\sin(\theta)\iff r^2 = 3r\sin(\theta)$. From my understanding,
$ r^2 = 3r\sin(\theta) \implies r = 0$ or $r = 3\sin(\theta)$ How can I show that $r = 0$ or $r = 3\sin(\theta)$ and $r = 3\sin(\theta)$ are equivalent statements?
My thoughts: For all values of $(r,\theta)$ that satisfy $r = 3\sin(\theta)$, $r = 3\sin(\theta)$ is true. This includes $(0,\pi),(3, \frac \pi2)$, etc. The ordered pair $(0,\pi)$ satisfies $r = 3\sin(\theta)$, but $r = 0$ itself does not.
I'm always unsure about multiplying both sides of an equation by a variable. Any help is appreciated. Thank you
When $r=0$ , $sin(\theta)=k\pi$
In either case, the converse will also hold and $r=0$ hence, you can multipy by $r$ since $r\neq0$ whenever $\theta \neq k\pi$