Solve the inequality $\dfrac{x-5}{x-1} \le -1$. Hence solve $\dfrac{\cos (x-4)}{\cos x} \le -1$ for $0 \le x \le \pi$.

1. Rearrange the inequality to form a single fraction comparing to zero.

2. Identify the critical values from the numerator and denominator.

$\text{Numerator: } 2x-6 = 0 \implies x = 3$
$\text{Denominator: } x-1 = 0 \implies x = 1$

3. Determine the sign in each interval to find the solution set, noting that the denominator cannot be zero.

$\text{Testing intervals around critical values:}$
$\text{For } x < 1, \text{ e.g. } x=0: \quad \frac{-6}{-1} = 6 > 0$
$\text{For } 1 < x < 3, \text{ e.g. } x=2: \quad \frac{-2}{1} = -2 < 0$
$\text{For } x > 3, \text{ e.g. } x=4: \quad \frac{2}{3} > 0$
$\text{Since we require } \le 0 \text{ and } x \neq 1 \text{:}$
$1 < x \le 3$