Wow, are you sure you have this right? Looks like something your professor made up while drunk. Assuming it is correct, rewrite it as:

\(\displaystyle r=8\cos^4 \theta [1 - \sqrt{(1- \cos \theta )/2}]^3+3\sin^2 \theta

\)

Multiplying both sides by \(\displaystyle r^{11/2}\) gives:

\(\displaystyle r^{13/2}=8r^4\cos^4 \theta [\sqrt r - \sqrt{(r- r\cos \theta )/2}]^3+3r^{7/2}r^2\sin^2 \theta

\)

Now substitute \(\displaystyle x=r\cos \theta, y=r\sin \theta,\) and \(\displaystyle r=\sqrt{x^2+y^2}.\)

Note: I took the posive square root in the half-angle identity \(\displaystyle \sin \theta/2 = \pm \sqrt{(1-\cos \theta)/2}.\) Convince yourself that this is justified.