Convert advanced parametric equation to regular/cartesian can anybody help me to convert following parametric equation in a form Y  =Y(X):
$$
x = cos(t) \sqrt{(2 - cos^2(3t))} \\
y = sin(t) \sqrt{(2 - cos^2(3t))}
$$
I've tried also with Wolfram Alpha and it seem not to work:
Reduce[x == Cos[t] Sqrt[2 - Cos[3 t]^2], {t}]

Actually what I need is a function in a form Y = Y(X) to get this kind a result:
http://postimg.org/image/6n8f839yp/
Best!
Vito

I would say
$x^2+y^2=(\cos^2t+\sin^2t) (2 - \cos^2(3t)) = 2 - \cos^2(3t)=1+\sin^2(3t)$
$\Rightarrow $equations in polar coordinates:  
$r = \sqrt{1+ \sin^2(3t)}=\cdots=\sqrt{1+ \sin^2t(3 \cos^2t-sin^2t)^2}$
and transformation from polar into Cartesian coordinates: 
$\displaystyle \,\cos t = \frac{x}{r},\,\,  \sin t= \frac{y}{r},\,\,r=\sqrt{x^2+y^2}$
$\displaystyle \Rightarrow r^2=1+\frac{y^2}{r^2}\left(\frac{3x^2}{r^2}-\frac{y^2}{r^2}\right)^2\Rightarrow \cdots \Rightarrow r^8 - r^6 - y^2 (3 x^2 - y^2)^2 = 0$
$\Rightarrow \,\,$Cartesian:
$(x^2+y^2)^4 - (x^2+y^2)^3 -y^2 (3 x^2 - y^2)^2=0$
$x^8 + 4x^6 y^2 - x^6 + 6x^4 y^4 - 12x^4 y^2 + 4x^2 y^6 + 3x^2 y^4 + y^8 - 2y^6 = 0$
See: Wolfram
