Eliminating $\theta$ from $x=\cos\theta(2-\cos 2\theta)$ and $y=\sin\theta(2-\sin 2\theta)$ 
If $$x=2\cos\theta-\cos\theta\cos 2\theta$$
$$y=2\sin\theta-\sin\theta\sin 2\theta$$
find a relation between $x$ and $y$ (not involving $\theta$).

Another trig elimination that has me stumped. One approach is to express as homogeneous functions,
$$x=3\sin^2\theta \cos\theta+\cos^3\theta$$
$$y=2\sin^3\theta+2\sin\theta\cos^2\theta-2\sin^2\theta\cos \theta$$
and by forming linear combinations of $x,y$ find some third power trig polynomials eg$(\cos \theta+\sin\theta)^3$. But I am not finding any success with this problem.
Note that this problem is from Hobson, Treatise on Plane Trigonometry 2 ed pg.97 #47. Slightly altered in the form I ask it. And may contain a typo, see discussion.

Let $u=\cos\theta$ so $x=3u-2u^3$ and $y=-2u\pm2\sqrt{1-u^2}+2u^3$.
Let $z=x+y$ so $4(1-u^2)=(y+2u-2u^3)^2=(z-u)^2$ giving $$5u^2-2zu+z^2-4=0\implies5u=z\pm2\sqrt{5-z^2}.$$ Apply the quadratic equality repeatedly to get \begin{align}x&=3u-2u\cdot5^{-1}(2zu-z^2+4)=(3+2\cdot5^{-1}z^2-8\cdot5^{-1})u-4\cdot5^{-1}zu^2\\&=(3+2\cdot5^{-1}z^2-8\cdot5^{-1})u-4\cdot5^{-2}z(2zu-z^2+4)\\&=(3+2\cdot5^{-1}z^2-8\cdot5^{-1}-8\cdot5^{-2}z^2)5^{-1}(z\pm2\sqrt{5-z^2})+4\cdot5^{-2}(z^2-4).\end{align} Simplification leads to $$125x=22(x+y)^3-45(x+y)\pm2(35+2(x+y)^2)\sqrt{5-(x+y)^2}.$$
