# How to solve a polar equation when $r$ is $r^2$ instead?

I have $r^2=-4\sinθ$

and also I'm asked to set $r=0$, after that locate θ. If I simply set $r^2=0$ after that I'll get $\sin(2θ)=0$. That does not appear right.

After that I'm asked to set $θ=0$ and afterwards locate $r$. If I make use of the $r^2=-4\sinθ$ and also set $θ=0$ after that I will certainly get "DNE". Not exactly sure what to do rather after that

EDIT:

Sorry every person I created the trouble below incorrect. It was intended to be r ^ 2 = - 4sin2θ

That is where the 2θ originated from

1
2022-06-07 14:31:08
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For any kind of actual number, $r=0$ if and also just if $r^2=0$, so "set [ting ] $r=0$" coincides as establishing $r^2$ to absolutely no. Equivalently: if $r=0$, after that $r^2=0$, so certainly you get that $r^2=0$.

Nonetheless, I do not recognize why you assume you get $\sin(2\theta)=0$. If $r^2=0$, after that $-4\sin(\theta)=0$. That suggests that $\sin(\theta)=0$ ; where did that $2$ originated from?

If you set $\theta=0$ rather, after that $\sin(\theta) = \sin(0)$. Just how much is $\sin(0)$? Just how much is that when increased by $-4$? And also what is the (just) value of $r$ that will make $r^2 = -4\sin(0)$ real?

Once more, I do not recognize why you assume you will certainly get "Does not exist" if you connect in $\theta=0$. This is merely not the instance. (Though, if you had $r^2 = -4\cos(\theta)$, and also searched for an actual value of $r$ for the instance $\theta=0$, after that you would certainly be incapable to locate one ; are you certain you are calculating $\sin(0)$ appropriately?)

3
2022-06-07 14:52:04
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1) It is right.

2) You do not get DNE, you get 0 (Sqrt [ - 4 *0 ] = 0)

It appears as a really wonderful number 8 loop. You appear to have actually recognized the inquiry, yet are doing not have some standard algebra/trig that the trouble was not made to examination.

0
2022-06-07 14:52:03
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