Convert rectangular equation $(x^2+y^2)^2 - 4(x^2-y^2) = 0$ to polar form So I am not sure how to do this problem : "Convert rectangular equation $(x^2+y^2)^2 - 4(x^2-y^2) = 0$ to polar form". Any help would be appreciated.

$$0=(x^2+y^2)^2 - 4(x^2-y^2) $$
Recall that $x^2+y^2=r^2$ and $x=r\cos(\theta)$ and $y=r\sin(\theta)$
$$  0=(r^2)^2-4(r^2 \cos^2(\theta)-r^2\sin^2(\theta))$$
$$ 0=r^4-4(r^2(\cos^2(\theta)-\sin^2(\theta))$$
$$ 0 = r^4-4r^2\cos(2\theta)$$
$$ 0 = r^2(r^2-4\cos(2\theta))$$
