Find Cartesian equation to $\space r = 2(1 + \sin(\frac{\theta}{2}))$

Question

Convert the polar equation to Cartesian: $\space r = 2(1 + \sin(\frac{\theta}{2})).$

I"ve tried substituting $\space \sin(\frac{\theta}{2}) \space$ to $\space \dfrac{\sqrt{1 + \cos\theta}}{2} \space$ and no result.

Answer 1

$$r=2(1+\sin\frac{\theta}{2}) \iff r-2=2\sin\frac{\theta}{2} \iff r-2=\sqrt{2(1-\cos \theta)} \iff (r-2)^2=2-2\cos\theta$$

$$\iff r^2-4r+4-2=-2\frac{x}{r} \iff x=-\frac{r}{2}(r^2-4r+2)$$

$$\iff x -2r^2 = - \frac{r}{2}(r^2+2) \iff (x -2r^2)^2 = \frac{r^2}{4}(r^2+2)^2$$

$$\iff x^2 -4xr^2+4r^4=\frac{r^2}{4}(r^4+4r^2+4)$$

$$\iff x^2 - 4x(x^2+y^2) + 4(x^2+y^2)^2 = \frac{(x^2+y^2)^3}{4}+(x^2+y^2)^2+x^2+y^2$$

$$\iff \boxed{3(x^2+y^2)^2-4x(x^2+y^2)=\frac{(x^2+y^2)^3}{4} + y^2}.$$

Answer 2

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

$2(r^2-4r+4) - 1=-\dfrac xr$

$2r^2-8r+7=-\dfrac xr$

$2r^3 - 8r^2 + 7r = -x$

$2r^3 + 7r = 8r^2 -x$

Now square both sides

$4r^6 + 49r^2 + 28r^4 = 64r^4 + x^2 - 16r^2x$

$4r^6 -36r^4 + r^2(49+16x) - x^2 = 0$

Substitute $r^2$ with $x^2 + y^2$.

$4(x^2+y^2)^3 -36(x^2+y^2)^2 + (x^2+y^2)(49+16x) - x^2 = 0$