Find $\sin 2 x, \cos 2 x$, and $\tan 2 x$ if $\cos x=-\frac{1}{\sqrt{10}}$ and $x$ terminates in quadrant III.
\[
\begin{array}{l}
\sin 2 x=\square \\
\cos 2 x=\square \\
\tan 2 x=\square
\end{array}
\]

Step 1 ：We are given that \(\cos x = -\frac{1}{\sqrt{10}}\) and that \(x\) is in quadrant III. In quadrant III, both sine and cosine are negative.

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

Step 3 ：Then we can use the double angle formulas \(\sin 2x = 2\sin x\cos x\), \(\cos 2x = \cos^2 x - \sin^2 x\), and \(\tan 2x = \frac{\sin 2x}{\cos 2x}\) to find \(\sin 2x\), \(\cos 2x\), and \(\tan 2x\).

Step 4 ：By substituting the given values, we find that \(\sin 2x = 0.6\), \(\cos 2x = -0.8\), and \(\tan 2x = -0.75\).

Step 5 ：So, the final answer is \(\sin 2 x=\boxed{0.6}\), \(\cos 2 x=\boxed{-0.8}\), and \(\tan 2 x=\boxed{-0.75}\).