All the permissible value of b ,a=sin(2x-b)if `a=0` and `x=S_(2)` is a subset of `(0, pi)` are given by

Options

A. `b in (-n pi, 2npi), n in Z`
B. `b in (-n pi, 2pi - npi), n in Z`
C. `b in (-n pi, npi), n in Z`
D. none of these

Explanation

To solve the problem, we need to find all permissible values of \( b \) such that \( a = \sin(2x - b) \) when \( a = 0 \) and \( x = S_2 \) is a subset of \( (0, \pi) \).

### Step-by-Step Solution:

1. **Set the Equation**: Since \( a = 0 \), we have:
\[ \sin(2x - b) = 0 \]

2. **Solve for \( 2x - b \)**: The sine function is zero at integer multiples of \( \pi \):
\[ 2x - b = n\pi \quad \text{where } n \in \mathbb{Z} \]

3. **Rearranging the Equation**: We can rearrange this to express \( x \):
\[ 2x = n\pi + b \]
\[ x = \frac{n\pi + b}{2} \]

4. **Subset Condition**: Given that \( x \) must be in the interval \( (0, \pi) \):
\[ 0 < \frac{n\pi + b}{2} < \pi \]

5. **Multiply by 2**: To eliminate the fraction, we multiply the entire inequality by 2:
\[ 0 < n\pi + b < 2\pi \]

6. **Separate the Inequalities**: This gives us two inequalities:
- \( n\pi + b > 0 \)
- \( n\pi + b < 2\pi \)

7. **Express \( b \)**: From the first inequality:
\[ b > -n\pi \]
From the second inequality:
\[ b < 2\pi - n\pi \]
This can be rewritten as:
\[ b < (2 - n)\pi \]

8. **Combine the Results**: Thus, we have:
\[ -n\pi < b < (2 - n)\pi \]

9. **Determine Values of \( n \)**: Since \( n \) is an integer, we can analyze the permissible values of \( b \) for different integer values of \( n \).

10. **Final Result**: The permissible values of \( b \) can be expressed as:
\[ b \in (-n\pi, (2-n)\pi) \quad \text{for } n \in \mathbb{Z} \]