This is a trigonometry problem that can be solved by using the **double-angle formulas** for sine, cosine and tangent. The double-angle formulas are:
$$\sin 2x = 2 \sin x \cos x$$
$$\cos 2x = \cos^2 x - \sin^2 x$$
$$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$$
To find $\sin 2x$, $\cos 2x$ and $\tan 2x$, we need to know the values of $\sin x$ and $\cos x$. We are given that $\cos x = \frac{4}{5}$, but we need to find $\sin x$. To do this, we can use the **Pythagorean identity**:
$$\sin^2 x + \cos^2 x = 1$$
Plugging in the given value of $\cos x$, we get:
$$\sin^2 x + \left(\frac{4}{5}\right)^2 = 1$$
Solving for $\sin x$, we get:
$$\sin x = \pm \frac{\sqrt{9}}{5} = \pm \frac{3}{5}$$
We need to choose the correct sign for $\sin x$ based on the **quadrant** where $x$ lies. Since we are not given any information about the angle $x$, we will assume that it is in the **first quadrant**, where both $\sin x$ and $\cos x$ are positive. Therefore, we choose the positive sign and get:
$$\sin x = \frac{3}{5}$$
Now, we can use the double-angle formulas to find $\sin 2x$, $\cos 2x$ and $\tan 2x$. Plugging in the values of $\sin x$ and $\cos x$, we get:
$$\sin 2x = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$$
$$\cos 2x = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{7}{25}$$
$$\tan 2x = \frac{2 \times \frac{3}{4}}{1 - \left(\frac{3}{4}\right)^2} = \frac{8}{7}$$
Therefore, the values of $\sin 2x$, $\cos 2x$ and $\tan 2x$ are:
$$\sin 2x = \frac{24}{25}$$
$$\cos 2x = \frac{7}{25}$$
$$\tan 2x = \frac{8}{7}$$
You can learn more about the double-angle formulas and the Pythagorean identity from the web search results that I found for you.