Sometimes functions are defined like `f(x)=max{sinx,cosx}`, then `f(x)` is splitted like `f(x)={{:(cosx, x in (0,(pi)/(4)]),(sinx, x in ((pi)/(4),(pi)/(2)]):}` etc.
If `f(x)=max{(1)/(2),sinx}`, then `f(x)=(1)/(2)` is defined when `x in `

Options

A. `((pi)/(6),pi)`
B. `((pi)/(6),(5pi)/(6))`
C. `[(pi)/(6),(5pi)/(6)]`
D. `[0,(pi)/(6)]`

Answer

To solve the problem where \( f(x) = \max\left\{\frac{1}{2}, \sin x\right\} \), we need to determine the intervals where \( f(x) = \frac{1}{2} \).

### Step-by-Step Solution:

1. **Understand the Function**:
The function \( f(x) \) takes the maximum value between \( \frac{1}{2} \) and \( \sin x \). This means that we need to find out when \( \sin x \) is less than or equal to \( \frac{1}{2} \).

2. **Set Up the Inequality**:
We need to solve the inequality:
\[ \sin x \leq \frac{1}{2} \]

3. **Find the Critical Points**:
The sine function equals \( \frac{1}{2} \) at specific angles:
\[ x = \frac{\pi}{6} \quad \text{and} \quad x = \frac{5\pi}{6} \]

4. **Analyze the Intervals**:
We will analyze the intervals around the critical points:
- From \( 0 \) to \( \frac{\pi}{6} \)
- From \( \frac{\pi}{6} \) to \( \frac{5\pi}{6} \)
- From \( \frac{5\pi}{6} \) to \( \pi \)

5. **Evaluate Each Interval**:
- **Interval \( (0, \frac{\pi}{6}) \)**: Here, \( \sin x \) is increasing from \( 0 \) to \( \frac{1}{2} \), so \( \sin x < \frac{1}{2} \).
- **Interval \( (\frac{\pi}{6}, \frac{5\pi}{6}) \)**: In this interval, \( \sin x \) increases to \( 1 \) at \( \frac{\pi}{2} \) and then decreases back to \( \frac{1}{2} \). Thus, \( \sin x > \frac{1}{2} \) for \( x \in \left(\frac{\pi}{6}, \frac{5\pi}{6}\right) \).
- **Interval \( (\frac{5\pi}{6}, \pi) \)**: Here, \( \sin x \) is decreasing and \( \sin x < \frac{1}{2} \).

6. **Combine the Results**:
From the analysis:
- \( f(x) = \frac{1}{2} \) when \( x \in [0, \frac{\pi}{6}] \) and \( x \in [\frac{5\pi}{6}, \pi] \).

### Final Answer:
Thus, \( f(x) = \frac{1}{2} \) is defined when:
\[ x \in [0, \frac{\pi}{6}] \cup [\frac{5\pi}{6}, \pi] \]