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.
 A: 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 $$125x=22(x+y)^3-45(x+y)\pm2(35+2(x+y)^2)\sqrt{5-(x+y)^2}.$$
A: Indulging the possibility/likelihood of a typographic error in the source material (which was the source of the problem in OP's previous question), consider the system
$$\begin{align}
x &= \cos\theta\, ( 2 - \cos2\theta ) \\
y &= \sin\theta\, ( 2 - \color{red}{\cos2\theta} ) \quad\color{red}{\leftarrow\text{instead of $\sin2\theta$}}
\end{align}$$
From here, we easily have, defining $r:=2-\cos2\theta$,
$$\begin{align}
x^2+y^2 &= (2-\cos2\theta)^2\phantom{\,\cos2\theta} = r^2 \\
x^2-y^2 &= (2-\cos2\theta)^2\cos2\theta  = r^2(2-r) = (x^2+y^2)(2-r)
\end{align}$$
whence $r(x^2+y^2) = x^2+3y^2$, so that

$$(x^2+y^2)^3 = ( x^2 + 3y^2 )^2 \tag{$\star$}$$

Alternatively, if we had
$$\begin{align}
x &= \cos\theta\, ( 2 - \color{red}{\sin2\theta} ) \quad\color{red}{\leftarrow\text{instead of $\cos2\theta$}}\\
y &= \sin\theta\, ( 2 - \sin2\theta ) 
\end{align}$$
then, with $r:=2-\sin2\theta$,
$$\begin{align}
x^2+y^2 &= (2-\sin2\theta)^2 \phantom{\sin2\theta\,} = r^2\\
2 x y &= ( 2 - \sin2\theta)^2\sin2\theta = (x^2+y^2)(2-r)
\end{align}$$
whence

$$(x^2 + y^2)^3 = 4 (x^2 - x y + y^2)^2 \tag{$\star\star$}$$

Of the two, I favor $(\star)$ as the correction of the ostensible typo. Either way, these solutions are more in keeping with the spirit of similar exercises in the source than the (perfectly valid) ones given in other answers to this question as stated.
A: Since there is a discussion over the possibility of typos in the source of the question and inspired by @Blue's answer, here's another (possible) typo fixed approach: [Note: I am presenting this answer because I was more concerned about the appearance of the question, not the answer :) ]
$$x=\cos\theta-\cos\theta\cos2\theta\tag1$$
$$y=\sin\theta-\sin\theta\sin2\theta\tag2$$
(Observe $x=\color{red}{\not2}\cos\theta-\cos\theta\cos2\theta$ and $y=\color{red}{\not2}\sin\theta-\sin\theta\sin2\theta$)
Adding two equations, $$x+y=\cos\theta+\sin\theta-\underbrace{(\cos\theta\cos2\theta+\sin\theta\sin2\theta)}_{\cos\theta}=\sin\theta$$
From the first equation we get, $$x=\cos\theta(1-\cos2\theta)=2\sin^2\theta\cos\theta$$
By squaring, $$x^2=4\sin^4\theta\cos^2\theta=4\sin^4\theta(1-\sin^2\theta)$$
Substituting the first result, $$x^2=4(x+y)^4(1-(x+y)^2)$$
