Find the range of `f(x)=(log)_([x-1])sinxdot`

Explanation

To find the range of the function \( f(x) = \log_{\lfloor x - 1 \rfloor}(\sin x) \), we will follow these steps:

### Step 1: Determine the domain of the function
The function \( f(x) \) is defined under certain conditions:
1. \( \sin x > 0 \)
2. \( \lfloor x - 1 \rfloor > 1 \) (since the base of the logarithm must be greater than 1)

### Step 2: Analyze \( \sin x \)
The sine function is positive in the intervals:
- \( (0, \pi) \)
- \( (2\pi, 3\pi) \)
- and so on.

### Step 3: Analyze \( \lfloor x - 1 \rfloor \)
For \( \lfloor x - 1 \rfloor > 1 \):
- This means \( x - 1 \geq 2 \) or \( x \geq 3 \).

### Step 4: Combine the conditions
The function \( f(x) \) is defined for \( x \geq 3 \) and within the intervals where \( \sin x > 0 \). Therefore, we need to consider:
- \( x \in [3, 4) \) (where \( \lfloor x - 1 \rfloor = 2 \))
- \( x \in [5, 6) \) (where \( \lfloor x - 1 \rfloor = 4 \))
- and so on.

### Step 5: Analyze the logarithm
Since \( \sin x \) varies between 0 and 1, we have:
- \( \log_{\lfloor x - 1 \rfloor}(\sin x) \) will approach \( -\infty \) as \( \sin x \) approaches 0 and will be 0 when \( \sin x = 1 \).

### Step 6: Determine the range of \( f(x) \)
- As \( \sin x \) approaches 0, \( f(x) \) approaches \( -\infty \).
- When \( \sin x = 1 \), \( f(x) = 0 \).

Thus, the range of \( f(x) \) is:
\[ (-\infty, 0] \]

### Final Answer
The range of \( f(x) = \log_{\lfloor x - 1 \rfloor}(\sin x) \) is \( (-\infty, 0] \).

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