We start with the equation given:
\[ 8 \tan x = 15 \]
From this, we can express \(\tan x\):
\[ \tan x = \frac{15}{8} \]
Now, recall that:
\[ \tan x = \frac{\sin x}{\cos x} \]
This allows us to rewrite the equation:
\[ \frac{\sin x}{\cos x} = \frac{15}{8} \]
By cross multiplying, we get:
\[ 8 \sin x = 15 \cos x \]
We can express \(\sin x\) in terms of \(\cos x\):
\[ \sin x = \frac{15}{8} \cos x \]
Now, we need to find the value of \( \sin x - \cos x \):
\[ \sin x - \cos x = \frac{15}{8} \cos x - \cos x \]
Factoring out \(\cos x\):
\[ \sin x - \cos x = \left( \frac{15}{8} - 1 \right) \cos x \]
Calculating \(\frac{15}{8} - 1\):
\[ \frac{15}{8} - 1 = \frac{15}{8} - \frac{8}{8} = \frac{15 - 8}{8} = \frac{7}{8} \]
Substituting back, we have:
\[ \sin x - \cos x = \frac{7}{8} \cos x \]
Next, we need to find \(\cos x\). We can use the identity:
\[ \sin^2 x + \cos^2 x = 1 \]
Substituting \(\sin x = \frac{15}{8} \cos x\):
\[ \left( \frac{15}{8} \cos x \right)^2 + \cos^2 x = 1 \]
Simplifying this:
\[ \frac{225}{64} \cos^2 x + \cos^2 x = 1 \]
Combining the terms:
\[ \left( \frac{225}{64} + \frac{64}{64} \right) \cos^2 x = 1 \]
\[ \frac{289}{64} \cos^2 x = 1 \]
Multiplying both sides by \(\frac{64}{289}\):
\[ \cos^2 x = \frac{64}{289} \]
Taking the square root:
\[ \cos x = \frac{8}{17} \quad \text{(we consider the positive value since $\tan x > 0$)} \]
Now substitute \(\cos x\) back into the expression for \(\sin x - \cos x\):
\[ \sin x - \cos x = \frac{7}{8} \cos x = \frac{7}{8} \cdot \frac{8}{17} = \frac{7}{17} \]
Thus, the value of \( \sin x - \cos x \) is:
\[ \boxed{\frac{7}{17}} \]