Given that $\tan x = \frac{6}{8}$ and $\pi < x < \frac{3\pi}{2}$, we need to find the value of $\cos 2x$.

1. Problem Description

Given that $\tan x = \frac{6}{8}$ and $\pi < x < \frac{3\pi}{2}$, we need to find the value of $\cos 2x$.

2. Solution Steps

First, we simplify the given $\tan x = \frac{6}{8} = \frac{3}{4}$.

Since $\pi < x < \frac{3\pi}{2}$, $x$ is in the third quadrant. In the third quadrant, both sine and cosine are negative.

We can use the identity $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$.

Substituting $\tan x = \frac{3}{4}$ into the formula, we get:

$\cos 2x = \frac{1 - (\frac{3}{4})^2}{1 + (\frac{3}{4})^2} = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{\frac{16-9}{16}}{\frac{16+9}{16}} = \frac{\frac{7}{16}}{\frac{25}{16}} = \frac{7}{16} \cdot \frac{16}{25} = \frac{7}{25}$.

Alternatively, consider a right triangle with legs of length 3 and

4. The hypotenuse is $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

Since $x$ is in the third quadrant, both $\sin x$ and $\cos x$ are negative. Therefore, $\sin x = -\frac{3}{5}$ and $\cos x = -\frac{4}{5}$.

We can use the formula $\cos 2x = \cos^2 x - \sin^2 x = (-\frac{4}{5})^2 - (-\frac{3}{5})^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$.

We can also use the formula $\cos 2x = 2\cos^2 x - 1 = 2(-\frac{4}{5})^2 - 1 = 2(\frac{16}{25}) - 1 = \frac{32}{25} - \frac{25}{25} = \frac{7}{25}$.

Or we can use the formula $\cos 2x = 1 - 2\sin^2 x = 1 - 2(-\frac{3}{5})^2 = 1 - 2(\frac{9}{25}) = 1 - \frac{18}{25} = \frac{25}{25} - \frac{18}{25} = \frac{7}{25}$.

3. Final Answer

The final answer is 7/

2

5.