$\eqalign{ & y = \sin (2t) \cr & x = \cos (t) \cr} $
therefore:
$\eqalign{ & \sin (t) = {y \over {2\cos t}} \cr as: & \cos (t) = x \cr & {\rm{ sin(t) = }}{y \over {2x}} \cr using: & {\sin ^2}t + {\cos ^2}t = 1 \cr & {\rm{ }}{x^2} + {\left( {{y \over {2x}}} \right)^2} = 1 \cr & {x^2} + {{{y^2}} \over {4{x^2}}} = 1 \cr} $
My question is, is there anything wrong with leaving it in this form? is there a "cartesian" form I must adhere to? The answer in the textbook decides to tackle it in a different manner and achieves a different result (which I am able to achieve through my method too), the answer in the textbook is:
$\eqalign{ & y = 2\sin (t)\cos (t) \cr & {\sin ^2}t + {x^2} = 1 \cr & {\sin ^2}t = 1 - {x^2} \cr & \sin t = \sqrt {1 - {x^2}} \cr & y = (2\sqrt {1 - {x^2}} )x \cr & y = 2x\sqrt {1 - {x^2}} \cr} $
Your solution is more correct in some sense. For instance, take $t=-\pi/3$. We get $$x=\cos(t) = 1/2 \text{ and } y=\sin(2t) = \sin(-2 \pi/3) = - \sqrt{3}/2,$$ which is admissible according to your equation $$x^2 + \dfrac{y^2}{4x^2} = 1$$ whereas the book solution for $x=1/2$ gives $y = \dfrac{\sqrt{3}}2$, which is not correct.