Answer
See image:
Work Step by Step
The curve seems to be sinusiodal, with period $\pi$ ...$ (\sin 2x)$,
moved to the right by $\displaystyle \frac{\pi}{2}\qquad...\qquad [\sin(2x-\frac{\pi}{2})]$
the horizontal axis is raised up by $1.5$ units $\displaystyle \qquad...\qquad [1.5+\sin(2x-\frac{\pi}{2})]$
The graph oscillates between 0.5 and 2, so we set amplitude =$\displaystyle \frac{2-0.5}{2}=0.75$
so we define $r=f(\theta))=1.5+0.75\displaystyle \sin(2\theta-\frac{\pi}{2})$
Now, build a table of values for $\theta=0,\pi/12,\pi/6,...$
(start with 0, add $\pi/12 $ to next $\theta$)
and for each $\theta$ calculate
x coordinate = $ f(\theta)\cdot\cos\theta,\qquad y=f(\theta)\cdot\sin\theta$
Plot the points and join with a smooth curve.
(Work done in geogebra.)
