#### Answer

$$\sin2x=\frac{4}{5}$$
$$\cos2x=-\frac{3}{5}$$

#### Work Step by Step

$$\tan x=2 \hspace{2cm}\cos x\gt0$$
$$\sin 2x=?\hspace{2cm}\cos2x=?$$
1) First, we need to calculate $\sin x$ and $\cos x$ as they are essential to the calculations of $\sin 2x$ and $\cos 2x$.
According to Pythagorean Identities:
$$\tan^2x+1=\sec^2x$$
According to Reciprocal Identities:
$$\sec x=\frac{1}{\cos x}$$
Therefore, $$\tan^2x+1=\frac{1}{\cos^2x}$$
$$\cos^2x=\frac{1}{\tan^2x+1}=\frac{1}{2^2+1}=\frac{1}{5}$$
$$\cos x=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}\hspace{1cm}(\cos x\gt0)$$
Now we find $\sin x$, using Quotient Identities:
$$\tan x=\frac{\sin x}{\cos x}$$
$$\sin x=\tan x\cos x=2\times\frac{\sqrt5}{5}=\frac{2\sqrt5}{5}$$
2) Now we can calculate $\sin2x$ and $\cos2x$ using Double-Angle Identities
$$\sin2x=2\sin x\cos x=2\times\frac{2\sqrt5}{5}\times\frac{\sqrt5}{5}=\frac{4}{5}$$
$$\cos 2x=\cos^2x-\sin^2x=\Big(\frac{\sqrt5}{5}\Big)^2-\Big(\frac{2\sqrt5}{5}\Big)^2=\frac{1}{5}-\frac{4}{5}=-\frac{3}{5}$$