Convert x^2+y^2=4y-2x to Polar Equation

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Homework Statement


Establish an equation in polar coordinates for the curve x^2+y^2=4y-2x


Homework Equations



n/a

The Attempt at a Solution


I know that x^2+y^2=r^2 so I used substitution, and now have r^2=4y-2x. Now this next part, I'm really not sure if I'm allowed to do this... i know that x=rcos@ (let @=theta) and y=rsin@. So i simply substituted to get r^2=4rsin@-2cos@. Is that allowed? Also, i feel like my answer should just look like r=... So should I take the sqaure root of both sides?
Thank you, very appreciated.
 
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You should have r^2 = 4rsin@ - 2rcos@. You lost an "r" on your cosine term. You can simplify by dividing both sides by r, but you don't want to take the square root of both sides. You normally have to be careful of dividing by r, in case r happens to be equal to zero. That's not a problem in this case, since the curve doesn't go through the origin.
 
Yeah, the missing r is a type. Dividing makes sense though, duh. thanks so much!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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