Range of `f(x)=(tan(pi[x^(2)-x]))/(1+sin(cosx))`

Answer

To find the range of the function \( f(x) = \frac{\tan(\pi [x^2 - x])}{1 + \sin(\cos x)} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Floor Function**:
The function contains the floor function, denoted as \([x^2 - x]\). The floor function returns the greatest integer less than or equal to \(x^2 - x\). The expression \(x^2 - x\) is a quadratic function, which can take any real value depending on \(x\).

**Hint**: Remember that the floor function outputs integers.

2. **Determine the Range of the Floor Function**:
Since \([x^2 - x]\) can take any integer value, we can denote it as \(n\), where \(n \in \mathbb{Z}\) (the set of all integers).

**Hint**: The quadratic \(x^2 - x\) can be analyzed by finding its vertex and roots to see its behavior.

3. **Evaluate the Numerator**:
The numerator becomes \(\tan(n\pi)\) where \(n\) is an integer. We know that \(\tan(n\pi) = 0\) for any integer \(n\) because \(\sin(n\pi) = 0\).

**Hint**: Recall that the tangent function is zero at integer multiples of \(\pi\).

4. **Evaluate the Denominator**:
The denominator is \(1 + \sin(\cos x)\). The function \(\sin(\cos x)\) oscillates between \(-1\) and \(1\) since \(\cos x\) ranges from \(-1\) to \(1\). Therefore, \(1 + \sin(\cos x)\) ranges from \(0\) to \(2\).

**Hint**: Consider the range of the sine function and how it affects the overall expression.

5. **Combine the Results**:
Since the numerator is \(0\), we have:
\[ f(x) = \frac{0}{1 + \sin(\cos x)} = 0 \]
for all \(x\) where the denominator is not zero. However, \(1 + \sin(\cos x)\) is never zero, as it ranges from \(0\) to \(2\).

**Hint**: Check if the denominator can ever be zero and confirm it cannot.

6. **Conclusion**:
Since \(f(x) = 0\) for all \(x\), the range of \(f(x)\) is simply the set containing the single value \(0\).

**Hint**: A constant function has a range that is just the constant value it outputs.

### Final Answer:
The range of \(f(x) = 0\).