If $\tan x=3$, then what is the value of ${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$? 
If $\tan x=3$, then what is the value of
  $${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$$

So what I did is change all the $\sin{2x}$ and $cos{2x}$ with double angle formulas, getting
$${3\cos^2{x}-3\sin^2{x}-4\sin{x}\cos{x}\over5\cos^2{x}-5\sin^2{x}+8\sin{x}\cos{x}}$$
Now I thought of changing the top part to $\sin{x}$ and bottom part to $\cos{x}$ hoping to somehow get $\tan{x}$ in this way, but I ultimately got just
$${3-6\sin^2{x}-4\sin{x}\cos{x}\over-5+10\cos^2{x}+8\sin{x}\cos{x}}$$
Had really no ideas what to either do after this, seems pretty unusable to me. Was there possibly a mistake I made in the transformation or maybe another way of solving this?

Since $\tan x = {\sin x \over \cos x}$ we have $\sin x = 3\cos x$ so
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}= {3\cos^2x-3\sin^2x-4\sin x \cos x\over 8\sin x \cos x +5\cos^2-5\sin^2 x}$$
$$ {3-27-12\over 24+5-45} = {9\over 4}$$ 
