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. **Understanding the Function**:
The function \( f(x) \) takes the maximum value between \( \frac{1}{2} \) and \( \sin x \). This means:
- If \( \sin x < \frac{1}{2} \), then \( f(x) = \frac{1}{2} \).
- If \( \sin x \geq \frac{1}{2} \), then \( f(x) = \sin x \).

2. **Finding Critical Points**:
We need to find the values of \( x \) for which \( \sin x = \frac{1}{2} \).
- The sine function equals \( \frac{1}{2} \) at \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \) within the interval \( [0, \pi] \).

3. **Analyzing the Intervals**:
- In the interval \( [0, \frac{\pi}{6}) \):
- \( \sin x < \frac{1}{2} \) (since \( \sin 0 = 0 \) and \( \sin \frac{\pi}{6} = \frac{1}{2} \)).
- Therefore, \( f(x) = \frac{1}{2} \).

- In the interval \( \left(\frac{\pi}{6}, \frac{5\pi}{6}\right) \):
- \( \sin x \) increases from \( \frac{1}{2} \) to \( 1 \) and then decreases back to \( \frac{1}{2} \).
- Therefore, \( f(x) = \sin x \) for \( x \in \left(\frac{\pi}{6}, \frac{5\pi}{6}\right) \).

- At \( x = \frac{5\pi}{6} \):
- \( \sin x = \frac{1}{2} \), so \( f(x) = \frac{1}{2} \).

- In the interval \( \left(\frac{5\pi}{6}, \pi\right) \):
- \( \sin x < \frac{1}{2} \) (as \( \sin \pi = 0 \)).
- Therefore, \( f(x) = \frac{1}{2} \).

4. **Conclusion**:
From the analysis, we find that \( f(x) = \frac{1}{2} \) in the intervals:
- \( [0, \frac{\pi}{6}] \) and \( [\frac{5\pi}{6}, \pi] \).

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