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?
 A: Since $\tan x = {\sin x \over \cos x}$ we have $\sin x = 3\cos x$ so
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}= {3\cos^2x-3\sin^2x-4\sin x \cos x\over 8\sin x \cos x +5\cos^2-5\sin^2 x}$$
$$ {3-27-12\over 24+5-45} = {9\over 4}$$ 
A: $$\tan2x=\dfrac{2\tan x}{1-\tan^2x}=-\dfrac34$$ then
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}=\dfrac{3-2\tan 2x}{4\tan2x+5}=\dfrac{9}{4}$$
A: The answer is $\displaystyle \frac 94$.
Alternative method.
I like this half angle identity: $\displaystyle \tan \frac 12 y = \frac{\sin y}{1 + \cos y}$
So $\displaystyle 3 = \tan x = \frac{\sin 2x}{1 + \cos 2x}$, giving $\displaystyle \sin 2x = 3 + 3\cos 2x$.
Substituting that into the original expression transforms it into:
$\displaystyle \frac{-3\cos 2x - 6}{17\cos 2x + 12}$
and using $\displaystyle \cos 2x = 2\cos^2x - 1$, that can be re-written:
$\displaystyle \frac{-6\cos^2x - 3}{34\cos^2x -5}$
Finally, going back to $\tan x = 3$, note that $\displaystyle 1+ \tan^2x = 10$, so $\displaystyle \sec^2x = 10$, so $\displaystyle \cos^2x = \frac 1{10}$.
Putting that into the expression yields the value $\displaystyle \frac 94$, which is the required answer.
