Polar to Cartesian: r = 3 + sin(theta/2)

Question

I am asked to convert the following polar function to cartesian: $$r = 3 + sin(\theta/2)$$

I would be able to do it if it weren't for the fraction.

I have already tried substituting the identity $sin(\theta/2) = \pm\sqrt{\frac{1 - cos(x)}{2}}$ but that is a dead end as far as I can tell.

Answer 1

$2(r-3)^2=1-\cos\theta$

$2r^2-12r-17=-\dfrac xr$

$x=r\{17+ 12r-2(x^2+y^2)\}$

$x-12(x^2+y^2)=r(17-2x^2-2y^2)$

Now square both sides