A special angle

Note that by the given information, $\sin{x} = 2 \cos{x}$ . By the double-angle formulas for each of the terms in our expression, we may reduce the desired expression to $2 \sin x \cos x + \cos^2 x - \sin^2 x + \frac{2 \tan x}{1 - \tan^2 x}$

Substituting $\tan x = 2$ and the derived equality, we have

$4 \cos ^2 x + \cos^2 x - 4 \cos^2 x - \frac{4}{3}$ $\cos^2 x - \frac{4}{3}$

Now for a bit of clever manipulation. We know that $\sin x = 2 \cos x$ , so squaring both sides and then adding $\cos^2 x$ yields $\sin^2 x + \cos^2 x = 5 \cos^2 x$ , so $1 = 5 \cos^2 x$ , so $\cos^2 x = \frac{1}{5}$ . Our desired value becomes

$\frac{1}{5} - \frac{4}{3}$

, which is $\frac{-17}{15}$ and our desired answer becomes $\boxed{32}$ .