Because the cotangent function has a divide by zero problem when #theta = 0 or pi#, this translates to the Cartesian restriction #y!=0#
Add #theta# to both sides:
#r = theta - 2cos^3(theta) - cot^2(theta)#
Write #cot^2(theta)# as #(cos^2(theta))/(sin^2(theta))#
#r = theta - 2cos^3(theta) - (cos^2(theta))/(sin^2(theta))#
Multiply both sides by #r^5sin^2(theta)#:
#r^6sin^2(theta) = thetar^3(r^2sin^2(theta)) - 2r^3cos^3(theta)r^2(sin^2(theta)) - r^5cos^2(theta)#
Substitute y for #rsin(theta)# and x for #rcos(theta)#
#r^4y^2 = thetar^3y^2 - 2x^3y^2 - r^3x^2; y!=0#
Use #(x^2 + y^2)^(1/2) = r# to make substitutions in accordance with the power of r:
#(x^2 + y^2)^(2)y^2 = theta(x^2 + y^2)^(3/2)y^2 - 2x^3y^2 - (x^2 + y^2)^(3/2)x^2; y!=0#
Collect like terms:
#(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)(thetay^2 - x^2) - 2x^3y^2; y!=0#
The substitution for #theta# breaks the above into 3 equations:
#(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)(tan^-1(y/x)y^2 - x^2) - 2x^3y^2; y > 0 and x > 0#
#(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)(tan^-1(y/x) + pi)y^2 - x^2) - 2x^3y^2; y != 0 and x < 0#
#(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)((tan^-1(y/x) + 2pi)y^2 - x^2) - 2x^3y^2; y < 0 and x > 0#
