If $\cos x=\frac{3}{5}, x$ in quadrant I, then find (without finding $x$ )
\[
\begin{array}{l}
\sin (2 x)= \\
\cos (2 x)= \\
\tan (2 x)=
\end{array}
\]

Step 1 ：We are given that \(\cos x = \frac{3}{5}\) and \(x\) is in quadrant I.

Step 2 ：We can use the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\) to find \(\sin x\).

Step 3 ：Substituting \(\cos x = \frac{3}{5}\) into the Pythagorean identity, we get \(\sin x = \frac{4}{5}\).

Step 4 ：We can use the double angle formulas to find \(\sin 2x\), \(\cos 2x\), and \(\tan 2x\).

Step 5 ：Using the formula \(\sin 2x = 2\sin x \cos x\), we substitute \(\sin x = \frac{4}{5}\) and \(\cos x = \frac{3}{5}\) to get \(\sin 2x = 0.96\).

Step 6 ：Using the formula \(\cos 2x = \cos^2 x - \sin^2 x\), we substitute \(\sin x = \frac{4}{5}\) and \(\cos x = \frac{3}{5}\) to get \(\cos 2x = -0.28\).

Step 7 ：Using the formula \(\tan 2x = \frac{\sin 2x}{\cos 2x}\), we substitute \(\sin 2x = 0.96\) and \(\cos 2x = -0.28\) to get \(\tan 2x = -3.43\).

Step 8 ：Final Answer: \(\sin (2 x)= \boxed{0.96}\), \(\cos (2 x)= \boxed{-0.28}\), \(\tan (2 x)= \boxed{-3.43}\)