Converting cartesian to polar form

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Given $$y^2 = x^2 - x^4,$$

how do you represent this in polar form?

I tried substituting $$x=r\cos \theta$$ and $$ y= r\cos \theta$$ which gave me $$r = \sqrt{\frac {\cos^2 \theta - \sin^2 \theta} {\cos^4 \theta}} $$

However, the square root does not give a real number e.g. when $$\theta=1$$ $r$ is undefined.

What would be the correct equation then?

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Perhaps a plot makes clear: Your equation defines a closed curve for $x\in(-1,1)$

   Plot[ {1, -1}*x Sqrt[(1 - x^2)], { x, -1, 1 }]

Lemniskate

Since |x| is bounded by 1, the replacement in polar coordinates yields

$$ \text{ParametricPlot[} \{\sin[ \phi], \cos[\phi] \sin[\phi\}, \{ \phi, -\pi, \pi \}]$$

same picture. The height of $1/2$ and the double loop is due to

$$\cos \phi\ \sin \phi = \frac{1}{2}\ \sin (2\phi)$$