Answer
$x^2+y^2=(3-2y)^2$
Work Step by Step
After multiplying with the denominator of the fraction:
$r(4+8\sin{\theta})=12\\4r+8r\sin{\theta}=12\\r+2r\sin{\theta}=3\\r=3-2r\sin{\theta}\\r^2=(3-2r\sin{\theta})^2$.
We know that $r^2=x^2+y^2$ and that $x=r\cos\theta,y=r\sin\theta$.
Hence $r^2=(3-2r\sin{\theta})^2$ becomes: $x^2+y^2=(3-2y)^2$
