Let's start with the given information:
[tex]\[
\sin (x) = -\frac{1}{3}
\][/tex]
and
[tex]\[
\cos (x) > 0
\][/tex]
Since [tex]\(\cos(x) > 0\)[/tex] and [tex]\(\sin(x) < 0\)[/tex], [tex]\(x\)[/tex] must be in the fourth quadrant where sine is negative and cosine is positive.
Using the Pythagorean identity:
[tex]\[
\sin^2(x) + \cos^2(x) = 1
\][/tex]
Substituting the given value of [tex]\(\sin(x)\)[/tex]:
[tex]\[
\left( -\frac{1}{3} \right)^2 + \cos^2(x) = 1
\][/tex]
[tex]\[
\frac{1}{9} + \cos^2(x) = 1
\][/tex]
Solving for [tex]\(\cos^2(x)\)[/tex]:
[tex]\[
\cos^2(x) = 1 - \frac{1}{9}
\][/tex]
[tex]\[
\cos^2(x) = \frac{9}{9} - \frac{1}{9}
\][/tex]
[tex]\[
\cos^2(x) = \frac{8}{9}
\][/tex]
Since [tex]\(\cos(x) > 0\)[/tex], we take the positive square root:
[tex]\[
\cos(x) = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}
\][/tex]
Next, we calculate [tex]\(\tan(x)\)[/tex]:
[tex]\[
\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}} = \frac{-1}{3} \cdot \frac{3}{2\sqrt{2}} = \frac{-1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4}
\][/tex]
Now, we use the double-angle formula for tangent:
[tex]\[
\tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)}
\][/tex]
Substituting [tex]\(\tan(x) = -\frac{\sqrt{2}}{4}\)[/tex]:
[tex]\[
\tan(2x) = \frac{2 \left( -\frac{\sqrt{2}}{4} \right)}{1 - \left( -\frac{\sqrt{2}}{4} \right)^2}
\][/tex]
[tex]\[
\tan(2x) = \frac{2 \left( -\frac{\sqrt{2}}{4} \right)}{1 - \frac{2}{16}}
\][/tex]
[tex]\[
\tan(2x) = \frac{-\frac{\sqrt{2}}{2}}{1 - \frac{1}{8}}
\][/tex]
[tex]\[
\tan(2x) = \frac{-\frac{\sqrt{2}}{2}}{\frac{8}{8} - \frac{1}{8}}
\][/tex]
[tex]\[
\tan(2x) = \frac{-\frac{\sqrt{2}}{2}}{\frac{7}{8}}
\][/tex]
[tex]\[
\tan(2x) = -\frac{\sqrt{2}}{2} \cdot \frac{8}{7} = -\frac{4\sqrt{2}}{7}
\][/tex]
Thus, the value of [tex]\(\tan(2x)\)[/tex] is:
[tex]\[
-\frac{4 \sqrt{2}}{7}
\][/tex]
So the correct answer is:
[tex]\[
-\frac{4 \sqrt{2}}{7}
\][/tex]
