If $\frac{\sin x}{\sin y}= {3}$ and $\frac{\cos x}{\cos y}= \frac{1}{2}$, then find $\frac{\sin2x}{\sin2y}+\frac{\cos2x}{\cos2y}$ 
Let $x$ and $y$ be real numbers such that $\dfrac{\sin x}{\sin y}= {3}$ and  $\dfrac{\cos x}{\cos y}= \dfrac{1}{2}$. Find a value of $$\dfrac{\sin2x}{\sin2y}+\dfrac{\cos2x}{\cos2y}.$$

My attempts:
Using the given condition and double angle formula, 
$\dfrac{\sin2x}{\sin2y} = \dfrac{3}{2}$
Now I am struggling to find the value of $\dfrac{\cos2x}{\cos2y}$
The simplest form I could reach was: 
$\dfrac{2\cos^2x-1}{8\cos^2x-1}$
How do I continue from here? Simpler methods to solve the problem are welcome.
PS: The answer is $\dfrac{49}{58}$  

$\sin{y}=3\sin{x}$ and $\cos{y}=2\cos{x}$.
Thus,
$$1=\sin^2y+\cos^2y=9\sin^2x+4\cos^2x=4+5\sin^2x,$$
which is impossible.
If $\sin{y}=\frac{1}{3}\sin{x}$ we obtain:
$$1=\sin^2y+\cos^2y=\frac{1}{9}\sin^2x+4\cos^2x=\frac{1}{9}+\frac{35}{9}\sin^2x.$$
Now, you can get $\sin^2x$, $\sin^2y$ and the rest for you. 
