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Answer 1 · 2017-03-10T12:12:32

#r=pm3/2sin(2theta)#

The pass equations are

#{(x=rcostheta),(y=rsintheta):}#

so

#(r^2)^3=3^2r^4cos^2thetasin^2theta# or

#r^2=3^2cos^2theta sin^2theta# or

#r=pm 3costheta sin theta# ---------(*)

using #sin(a+b)=sin acos b+cosa sinb# and making #a=b#

we get at

#sin(2a)=2sina cosa# then #sin theta cos theta = 1/2 sin(2theta)#

substituting into (*) we obtain

#r=pm3/2sin(2theta)#

Attached a plot.

enter image source here