Put in x=r\cos(\theta), y=r\sin(\theta)x=rcos(θ),y=rsin(θ). Then
r^2(\cos^2(\theta)-\sin^2\theta))=6r\cos(\theta)\pm r\sin(\theta)+r^2\cos^2(\theta)\sin(\theta)r2(cos2(θ)−sin2θ))=6rcos(θ)±rsin(θ)+r2cos2(θ)sin(θ)
Since every term has a factor of rr, either r=0r=0 or you can divide by rr. The latter gives
r(\cos^2(\theta)-\sin^2\theta))=6\cos(\theta)\pm \sin(\theta)+r^2\cos^2(\theta)\sin(\theta)r(cos2(θ)−sin2θ))=6cos(θ)±sin(θ)+r2cos2(θ)sin(θ)
This already contains r=0r=0 when \tan(\theta)=\pm 6tan(θ)=±6 and so it's the complete solution.
