Convert the system in the polar coordinates r'=(1-r), theta' =sin(theta/2)^2 into cartesian coordinates.
You have a linear transformation from one polar coordinate system to another: $\displaystyle r'= 1- r$ and $\displaystyle \theta'= sin^2(\theta/2)$. I would start by using the "half angle formula":
$\displaystyle sin(\theta/2)= \sqrt{(1/2)(1- cos(\theta)}$ so that $\displaystyle \theta'= sin^2(\theta/2)= \frac{1}{2}(1- cos(\theta)$. Now use the fact that $\displaystyle \theta= arctan\left(\frac{y}{x}\right)$, $\displaystyle r= \sqrt{x^2+ y^2}$, $\displaystyle \theta'= arctan\left(\frac{y'}{x'}\right)$, and $\displaystyle r'= \sqrt{x'^2+ y'^2}$