#### Answer

$\displaystyle \cos 2x=-\frac{3}{5}$
$\displaystyle \sin 2x=\frac{4}{5}$

#### Work Step by Step

$\tan x$ and $\cos x$ are positive $\Rightarrow \sin x$ is positive, since $\displaystyle \tan x=\frac{\sin x}{\cos x}$.
This also means that x is in quadrant I,
$0^{\mathrm{o}} < x < 90^{\mathrm{o}}\qquad/\times 2$
$0^{\mathrm{o}} < 2x < 180^{\mathrm{o}}, $
.. so 2x is either in quadrant I or II (*)
Double-Angle Identity for tan:
$\displaystyle \tan 2x=\frac{2\tan x}{1-\tan^{2}x}=\frac{2\cdot 2}{1-2^{2}}=-\frac{4}{3}$
$\tan 2x $is negative, so one of $\sin 2x,\ \cos 2x$ is negative, and the other is positive.
With (*), 2x is in quadrant II, where its sine is positive and cosine negative.
Pythagorean Identity:
$\sec^{2}2x=1+\tan^{2}2x=$
$=1+(-\displaystyle \frac{4}{3})=1+\frac{16}{9}=\frac{25}{9}$,
... cosine and secant are reciprocals, so
$\displaystyle \cos^{2}2x=\frac{9}{25}$ and
$\displaystyle \cos 2x=-\frac{3}{5}$ (cos2x is negative)
Pythagorean Identity:
$\sin^{2}\theta+\cos^{2}\theta=1,$
$\displaystyle \sin 2x=\sqrt{1-(-\frac{3}{5})^{2}}=\sqrt{\frac{25-9}{25}}=\frac{4}{5}$