Find $\sin (2x),\cos (2x)$, and $\tan (2x)$ from the given information.
$$\begin{array}{l}\tan (x)=-\frac{1}{6},\cos (x)>0\\ \sin (2x)=◻\\ \cos (2x)=◻\\ \tan (2x)=◻\end{array}$$
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Step 1 ：We are given that $\tan (x)=-\frac{1}{6}$ and $\cos (x)>0$.

Step 2 ：We need to find $\sin (2x)$, $\cos (2x)$, and $\tan (2x)$.

Step 3 ：We know that $\sin (2x)=2\sin (x)\cos (x)$, $\cos (2x)={\cos }^2(x)-{\sin }^2(x)$, and $\tan (2x)=\frac{2\tan (x)}{1-{\tan }^2(x)}$.

Step 4 ：We can find $\sin (x)$ and $\cos (x)$ from the given $\tan (x)$ using the Pythagorean identity ${\sin }^2(x)+{\cos }^2(x)=1$ and the fact that $\tan (x)=\frac{\sin (x)}{\cos (x)}$.

Step 5 ：Since $\cos (x)>0$ and $\tan (x)<0$, we know that $\sin (x)<0$ (because the tangent of an angle is negative if and only if the sine and cosine have opposite signs).

Step 6 ：So, we can find $\sin (x)$ and $\cos (x)$, and then substitute these values into the formulas for $\sin (2x)$, $\cos (2x)$, and $\tan (2x)$ to find the answers.

Step 7 ：Final Answer: $$\begin{array}{l}\sin (2x)=\boxed{{\displaystyle -0.324}}\\ \cos (2x)=\boxed{{\displaystyle 0.946}}\\ \tan (2x)=\boxed{{\displaystyle -0.343}}\end{array}$$