would you sketch the curve with the polar equation r = 2cos(2theta), and express the equation in rectangular coordinates, that is what must be represented.
$\displaystyle r = 2 \cos2\theta$
$\displaystyle r^2 = x^2 + y^2$
...for $\displaystyle r>0~,~~r = + \sqrt{x^2+y^2}$
...for $\displaystyle r<0~,~~r = - \sqrt{x^2+y^2}$
$\displaystyle \tan\theta = \frac{y}{x}$
...for $\displaystyle -\frac{\pi}{2}<\theta<\frac{\pi}{2}~,~~\theta = \arctan(\frac{y}{x})$
...for $\displaystyle \frac{\pi}{2}<\theta<\frac{3\pi}{2}~,~~\theta = \arctan(\frac{y}{x})+\pi$
Well, if we look carefully at $\displaystyle \theta$..
$\displaystyle \theta = \arctan(\frac{y}{x})$
$\displaystyle \theta = \arctan(\frac{y}{x})+\pi$
$\displaystyle r = 2 \cos(2\theta)$
$\displaystyle r = 2 \cos(2(\arctan(\frac{y}{x})+\pi))$
$\displaystyle r = 2 \cos(2\arctan(\frac{y}{x})+2\pi))$
which is equal to $\displaystyle r = 2 \cos2(\arctan(\frac{y}{x}))$
That means, $\displaystyle \theta = \arctan(\frac{y}{x})$ and $\displaystyle \theta = \arctan(\frac{y}{x})+\pi$ are the same. Then, using one of them is enough.
So, we have
$\displaystyle r = 2 \cos2\theta$
...for $\displaystyle r>0~,~~r = + \sqrt{x^2+y^2}$
...for $\displaystyle r<0~,~~r = - \sqrt{x^2+y^2}$
$\displaystyle \theta = \arctan(\frac{y}{x})$
Placing them will give us two functions:
$\displaystyle \sqrt{x^2+y^2} = 2\cos(2\arctan(\frac{y}{x}))$
$\displaystyle -\sqrt{x^2+y^2} = 2\cos(2\arctan(\frac{y}{x}))$