How to graph r 2 = cos ( 2 θ ) ?

the first and fourth quadrants ${\displaystyle \left|\theta \right|\le \frac{\pi }{4}}$ In the 2nd and 3rd,
${\displaystyle |\pi -\theta |\le \frac{\pi }{4}}$.
I used infinity symbol ${\displaystyle \infty }$ to depict the shape of this double-
loop, looking like a fallen 8. For getting 8-erect, the equation is
${\displaystyle r^2=-\cos 2\theta }$
Use a table
${\displaystyle \left\{(r,\theta )\right\}.=\{(0,-\frac{\pi }{4})(\frac{1}{\sqrt{2}},-\frac{\pi }{8})(1,0)(\frac{1}{\sqrt{2}},\frac{\pi }{8})(0,}$
${\displaystyle \frac{\pi }{4}\left)\right\}}$, for one loop. Its mirror image with respect to ${\displaystyle \theta =\frac{\pi }{2}}$
is the other loop.
Strictly, ${\displaystyle r=\sqrt{x^2+y^2}\ge 0}$.
Graphs of both ${\displaystyle r^2=\pm \cos 2\theta }$ are combined.