<s>[INST] 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?
 [/INST] $$\tan2x=\dfrac{2\tan x}{1-\tan^2x}=-\dfrac34$$ then
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}=\dfrac{3-2\tan 2x}{4\tan2x+5}=\dfrac{9}{4}$$
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