How to solve a polar equation when $r$ is $r^2$ instead? I have $r^2=-4\sinθ$
and I'm asked to set $r=0$, then find θ. If I just set $r^2=0$ then I'll get $\sin(2θ)=0$. That doesn't seem right. 
Then I'm asked to set $θ=0$ and then find $r$. If I use the $r^2=-4\sinθ$ and set $θ=0$ then I will get "DNE". Not sure what to do instead then...
EDIT:
Sorry everyone I wrote the problem here wrong. It was supposed to be r^2=-4sin2θ
That's where the 2θ came from
 A: For any real number, $r=0$ if and only if $r^2=0$, so "set[ting] $r=0$" is the same as setting $r^2$ to zero. Equivalently: if $r=0$, then $r^2=0$, so of course you get that $r^2=0$.
However, I don't understand why you think you get $\sin(2\theta)=0$. If $r^2=0$, then $-4\sin(\theta)=0$. That means that $\sin(\theta)=0$; where did that $2$ come from?
If you set $\theta=0$ instead, then $\sin(\theta) = \sin(0)$. How much is $\sin(0)$? How much is that when multiplied by $-4$? And what is the (only) value of $r$ that will make $r^2 = -4\sin(0)$ true? 
Again, I don't understand why you think you will get "Does not exist" if you plug in $\theta=0$. This is simply not the case. (Though, if you had $r^2 = -4\cos(\theta)$, and tried to find a real value of $r$ for the case $\theta=0$, then you would be unable to find one; are you sure you are computing $\sin(0)$ correctly?)
A: 1) It is right.
2) You do not get DNE, you get 0... (Sqrt[-4*0] = 0)
It comes out as a very nice figure 8 loop.
You seem to have understood the question, but are lacking some basic algebra/trig that the problem was not designed to test.
