Find the domain and range of f(x)=`log(1/sqrt([cosx]-[sinx]))`. [.] denotes the greatest integar function.

Answer

To find the domain and range of the function \( f(x) = \log\left(\frac{1}{\sqrt{[\cos x] - [\sin x]}}\right) \), where \([.]\) denotes the greatest integer function, we need to ensure that the argument of the logarithm is positive.

### Step 1: Determine the condition for the logarithm
The argument of the logarithm must be greater than zero:
\[ \frac{1}{\sqrt{[\cos x] - [\sin x]}} > 0 \]
This implies:
\[ [\cos x] - [\sin x] > 0 \]

### Step 2: Analyze the greatest integer function
The greatest integer function \([y]\) gives the largest integer less than or equal to \(y\). Therefore, we need to analyze the values of \([\cos x]\) and \([\sin x]\).

### Step 3: Identify ranges for \(\cos x\) and \(\sin x\)
The values of \(\cos x\) and \(\sin x\) oscillate between -1 and 1. Thus:
- \([\cos x]\) can be -1 or 0.
- \([\sin x]\) can be -1 or 0.

### Step 4: Determine when \([\cos x] - [\sin x] > 0\)
We will analyze the possible cases for \([\cos x]\) and \([\sin x]\):
1. If \([\cos x] = 0\) and \([\sin x] = 0\), then \([\cos x] - [\sin x] = 0\) (not valid).
2. If \([\cos x] = 0\) and \([\sin x] = -1\), then \([\cos x] - [\sin x] = 1\) (valid).
3. If \([\cos x] = -1\) and \([\sin x] = 0\), then \([\cos x] - [\sin x] = -1\) (not valid).
4. If \([\cos x] = -1\) and \([\sin x] = -1\), then \([\cos x] - [\sin x] = 0\) (not valid).

From this analysis, the only valid case is when \([\cos x] = 0\) and \([\sin x] = -1\).

### Step 5: Determine the intervals for \(x\)
- \([\cos x] = 0\) occurs when \(0 \leq \cos x < 1\), which corresponds to intervals where \(x\) is in the form of \((2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2})\) for integers \(n\).
- \([\sin x] = -1\) occurs when \(-1 \leq \sin x < 0\), which corresponds to intervals where \(x\) is in the form of \((2n\pi - \frac{3\pi}{2}, 2n\pi - \frac{\pi}{2})\) for integers \(n\).

### Step 6: Find the intersection of intervals
The valid intervals for \(x\) where both conditions are satisfied must be found. The intersection of these intervals leads to:
- For \(n = 0\): The intersection occurs in the interval \((-\frac{\pi}{2}, 0)\) and \((-\frac{3\pi}{2}, -\pi)\).
- For \(n = 1\): The intersection occurs in the interval \((\frac{\pi}{2}, \pi)\) and \((\frac{\pi}{2}, \frac{3\pi}{2})\).

### Domain and Range
Thus, the domain of \(f(x)\) can be expressed as:
\[ x \in \left(-\frac{\pi}{2}, 0\right) \cup \left(\frac{3\pi}{2}, 2\pi\right) \]

The range of \(f(x)\) is determined by the values of \(f(x)\) as \(x\) varies in the domain. Since the logarithm function approaches negative infinity as its argument approaches zero, and since the argument is positive in the defined intervals, the range is:
\[ (-\infty, 0) \]

### Final Answer
- **Domain**: \( x \in \left(-\frac{\pi}{2}, 0\right) \cup \left(\frac{3\pi}{2}, 2\pi\right) \)
- **Range**: \( (-\infty, 0) \)