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 \(\frac{\cos x - 1}{\tan x} > 0\), we start by analyzing the components. The term \(\cos x - 1\) is zero when \(x = 0\) and \(x = 2\pi\) and is negative for \(0 < x < 2\pi\). Therefore, this term contributes a negative value throughout the interval except at the endpoints. Next, we examine \(\tan x\). The tangent function is positive in the intervals \((0, \frac{\pi}{2})\) and \((\pi, \frac{3\pi}{2})\) and negative in \((\frac{\pi}{2}, \pi)\) and \((\frac{3\pi}{2}, 2\pi)\). Now, for the fraction \(\frac{\cos x - 1}{\tan x}\) to be positive, both the numerator and the denominator must have the same sign. The numerator \(\cos x - 1\) is non-positive (zero at \(0\) and \(2\pi\) and negative elsewhere), and \(\tan x\) is positive in the intervals stated above. Thus, the solution is in the interval where \(\tan x\) is negative while \(\cos x - 1\) is also negative which occurs in: - The interval \( (0, \frac{\pi}{2})\), where \( \tan x\) is positive - The interval \( ( \frac{\pi}{2}, \pi)\), where \(\tan x\) is negative - The interval \( ( \frac{3\pi}{2}, 2\pi)\), where \(\tan x\) is positive again However, we need the intervals where both numerator and denominator are negative or both are positive, which gives us the intervals \( ( \frac{\pi}{2}, \pi)\). Thus, the final answer in interval notation is: \[ \boxed{\left( \frac{\pi}{2}, \pi \right)} \]