Find the values of x on \( 0 \leq x \leq 2 \pi \) that satisfy the inequality. Write your answer in interval notation. \( \frac{\cos x-1}{\tan x}>0 \)

To solve the inequality: \[ \frac{\cos x - 1}{\tan x} > 0 \quad \text{for} \quad 0 \leq x \leq 2\pi, \] let's analyze the expression step by step. ### Step 1: Simplify the Inequality First, recall that \(\tan x = \frac{\sin x}{\cos x}\). Substituting this in, we get: \[ \frac{\cos x - 1}{\tan x} = \frac{\cos x - 1}{\frac{\sin x}{\cos x}} = \frac{(\cos x - 1) \cos x}{\sin x} \] However, analyzing the original form \(\frac{\cos x - 1}{\tan x}\) is more straightforward. ### Step 2: Determine the Signs of \(\cos x - 1\) and \(\tan x\) 1. **\(\cos x - 1\):** - \(\cos x \leq 1\) for all \(x\). - \(\cos x - 1 = 0\) when \(x = 0\) or \(x = 2\pi\). - Therefore, \(\cos x - 1 \leq 0\) for all \(x\) in \([0, 2\pi]\), with equality only at \(x = 0\) and \(x = 2\pi\). 2. **\(\tan x\):** - \(\tan x\) is positive in the **first** and **third** quadrants: \( (0, \frac{\pi}{2}) \) and \( (\pi, \frac{3\pi}{2}) \). - \(\tan x\) is negative in the **second** and **fourth** quadrants: \( (\frac{\pi}{2}, \pi) \) and \( (\frac{3\pi}{2}, 2\pi) \). - \(\tan x\) is undefined at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\). ### Step 3: Analyze the Inequality \(\frac{\cos x - 1}{\tan x} > 0\) For the fraction to be positive: - Since \(\cos x - 1 \leq 0\), the fraction \(\frac{\cos x - 1}{\tan x}\) is positive **only when** \(\tan x\) is negative. Thus, the inequality holds when: \[ \tan x < 0 \] This occurs in the **second** and **fourth** quadrants: - \( \left(\frac{\pi}{2}, \pi\right) \) - \( \left(\frac{3\pi}{2}, 2\pi\right) \) ### Conclusion The values of \(x\) in the interval \([0, 2\pi]\) that satisfy the inequality \(\frac{\cos x - 1}{\tan x} > 0\) are: \[ \left( \frac{\pi}{2}, \pi \right) \cup \left( \frac{3\pi}{2}, 2\pi \right) \] **Answer:** \(\left( \dfrac{\pi}{2},\; \pi\, \right) \cup \left( \dfrac{3\,\pi}{2},\; 2\,\pi \right)\)