Is it possible to convert the polar equation $\ r = k \cos (\theta n) + 2$ into cartesian form? 
Is it possible to convert the polaer equation 
  $$\ r =  k \cos (\theta n) + 2$$ 
  into cartesian form? Here, $k$ is some constant and $n$ is any positive whole number greater than $2$.

The farthest that I managed to get was:
$$r^2 = kr \cos(\theta n) +2r$$
$$\to\qquad r^2 - 2r - kr\cos(\theta n) = 0$$
$$\to \qquad x^2 +y^2 - 2 \sqrt{x^2 +y^2} -kr \cos (\theta n) = 0$$
I noticed that $\cos(\theta n)$ is reminiscent of the Chebychev polynomials and figured that $r\cos(\theta n)$ could also be generalized into polyomials in terms of $x$, but that turns out to be impossible.

If you want an expression that holds for all $n \geq 2$ then it might be a tad ugly in Cartesian, because:
$$\cos(n\theta) = \sum_{\text{even }k} (-1)^{k/2}{n \choose k}\cos^{n-k} \theta \sin^k \theta \\ = (x^2+y^2)^{-n/2}\sum_{\text{even }k} (-1)^{k/2}{n \choose k} x^{n-k}y^k.$$
But this will get you there.
