I have $r^2=2\cos2\theta$ and I'm being asked to convert this equation to rectangular coordinates. So I'm using double angle trigonometric identities to get:
$$ r^2=2\cos2\theta \\ r^2=2(\cos^2\theta-\sin^2\theta) \\ r^2=2\cos^2\theta-2\sin^2\theta $$
Then I do the following substitutions based on the facts that $y=r\sin\theta$ and $x=r\cos\theta$:
$$ r^2=\frac{2x^2}{r^2}-\frac{2y^2}{r^2} \\ (r^2)^2=2x^2-2y^2 \\ (x^2+y^2)^2=2x^2-2y^2 $$
However this doesn't feel right, since I would be getting a fourth-degree equation. Can you find any error in my development? Or how should I go on solving this problem? Thanks in advance.
In order to plot this curve, first consider the equation
$$r^2 = 2\cos 2\theta$$
Notice that the left hand side is always positive, so that induces a domain restriction on $\theta$.
$$|2\theta| \leq \frac{\pi}{2} \implies -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$$
Then notice that there is a symmetry $\theta \mapsto -\theta$, which means that the curve is symmetric about the $x$ axis (which is the ray $\theta = 0$), so we only need to know what the curve looks like on the top half and we can reflect it down to get the rest.
From here, it is best to pick example points and tabulate them:
$$\begin{array}{c | c} \theta & r \\ \hline 0 & \pm\sqrt{2} \\ \frac{\pi}{12} & \pm \sqrt[4]{3} \\ \frac{\pi}{8} & \pm \sqrt[4]{2} \\ \frac{\pi}{6} & \pm 1 \\ \frac{\pi}{4} & 0 \\ \end{array} $$
Plot the points and connect them as smoothly as you can, then reflect it across the $x$ axis. Drawing this out, we get a familiar shape, a lemniscate (or infinity symbol) figure.
One can confirm this is indeed a lemniscate by looking at the equation you derived. It is in the correct mathematical form for a lemniscate, with $c=1$.