Given that $\cos x = -\frac{4}{5}$ and $\tan x > 0$, find

1. Determine the quadrant of $x$ and find $\sin x$

Since $\cos x = -\frac{4}{5} < 0$ and $\tan x > 0$, $x$ lies in the 3rd quadrant.
In the 3rd quadrant, $\sin x < 0$.
Using $\sin^2 x + \cos^2 x = 1$:
$\sin^2 x = 1 - \left(-\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}$
$\sin x = -\frac{3}{5}$

2. Calculate $\sin 2x$ using the double angle formula

$\sin 2x = 2 \sin x \cos x$
$\sin 2x = 2 \left(-\frac{3}{5}\right) \left(-\frac{4}{5}\right)$
$\sin 2x = \frac{24}{25}$

1. Determine the quadrant of $x$ and deduce the quadrant of $\frac{x}{2}$

Since $\cos x = -\frac{4}{5} < 0$ and $\tan x > 0$, $x$ must be in the third quadrant.
Thus, $180^\circ < x < 270^\circ$.
Dividing by 2 gives $90^\circ < \frac{x}{2} < 135^\circ$.
Therefore, $\frac{x}{2}$ is in the second quadrant, where cosine is negative.

2. Use the double-angle formula to relate $\cos x$ to $\cos \frac{x}{2}$

$\cos x = 2\cos^2 \frac{x}{2} - 1$
$-\frac{4}{5} = 2\cos^2 \frac{x}{2} - 1$

3. Solve for $\cos \frac{x}{2}$ and select the correct sign

$2\cos^2 \frac{x}{2} = 1 - \frac{4}{5} = \frac{1}{5}$
$\cos^2 \frac{x}{2} = \frac{1}{10}$
$\cos \frac{x}{2} = \pm\frac{1}{\sqrt{10}}$
Since $\frac{x}{2}$ is in the second quadrant, $\cos \frac{x}{2} < 0$.
$\cos \frac{x}{2} = -\frac{1}{\sqrt{10}}$