I have attacked this equation from many sides and I can't figure out how to get 'r' by itself in the rectangular to polar conversion.
$$\ x^3+xy+y^2=5 $$ What makes this problem difficult is the 'xy' and 'x^3' terms making it difficult to simplify once the x's and y's are substituted with sin's and cosines'
I have a feeling there is a substitution that I could do somewhere because otherwise, I can't separate the r's and theta's
This is the closest I've gotten:
subtracting x^3 from both sides: $$\ xy+y^2=5-x^3 $$
convert: $$\ r^2cos(\theta)sin(\theta)+r^2sin^2(\theta)=5-r^3cos^3(\theta) $$
divide both sides by r^2: $$\ cos(\theta)sin(\theta)+sin^2(\theta)= (5/r^2)-rcos^3(\theta) $$
As @Andrei commented, I suppose a typo and that $x^3$could be $x^2$ instead.
Otherwise, solving the cubic equation in $r$, the discriminant would be $$\Delta=20 \sin ^3(\theta )\left(\sin (\theta )+ \cos (\theta )\right)^3-675 \cos ^6(\theta )$$ which, for the range $0 \leq \theta \leq 2\pi$, would be positive (then three real roots in $r$) for $$0.938845 < \theta < 1.97069$$