Because the cotangent function has a divide by zero problem when theta = 0 or piθ=0orπ, this translates to the Cartesian restriction y!=0y≠0
Add thetaθ to both sides:
r = theta - 2cos^3(theta) - cot^2(theta)r=θ−2cos3(θ)−cot2(θ)
Write cot^2(theta)cot2(θ) as (cos^2(theta))/(sin^2(theta))cos2(θ)sin2(θ)
r = theta - 2cos^3(theta) - (cos^2(theta))/(sin^2(theta))r=θ−2cos3(θ)−cos2(θ)sin2(θ)
Multiply both sides by r^5sin^2(theta)r5sin2(θ):
r^6sin^2(theta) = thetar^3(r^2sin^2(theta)) - 2r^3cos^3(theta)r^2(sin^2(theta)) - r^5cos^2(theta)r6sin2(θ)=θr3(r2sin2(θ))−2r3cos3(θ)r2(sin2(θ))−r5cos2(θ)
Substitute y for rsin(theta)rsin(θ) and x for rcos(theta)rcos(θ)
r^4y^2 = thetar^3y^2 - 2x^3y^2 - r^3x^2; y!=0r4y2=θr3y2−2x3y2−r3x2;y≠0
Use (x^2 + y^2)^(1/2) = r(x2+y2)12=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(x2+y2)2y2=θ(x2+y2)32y2−2x3y2−(x2+y2)32x2;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(x2+y2)2y2=(x2+y2)32(θy2−x2)−2x3y2;y≠0
The substitution for thetaθ 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(x2+y2)2y2=(x2+y2)32(tan−1(yx)y2−x2)−2x3y2;y>0andx>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(x2+y2)2y2=(x2+y2)32(tan−1(yx)+π)y2−x2)−2x3y2;y≠0andx<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(x2+y2)2y2=(x2+y2)32((tan−1(yx)+2π)y2−x2)−2x3y2;y<0andx>0
