One root of the equation `cos x-x+1/2=0` lies in the interval (A) `[0,pi/2]` (B) `[-pi/2,0]` (C) `[pi/2,0]` (D) none

Options

A. `(0, pi/2)`
B. `(- pi/2, 0)`
C. `(pi/2, pi)`
D. `(pi, (3pi)/2)`

Explanation

To solve the equation \( \cos x - x + \frac{1}{2} = 0 \) and determine in which interval one of its roots lies, we will analyze the function step by step.

### Step 1: Define the function
Let \( f(x) = \cos x - x + \frac{1}{2} \).

### Step 2: Evaluate \( f(0) \)
Calculate \( f(0) \):
\[ f(0) = \cos(0) - 0 + \frac{1}{2} = 1 - 0 + \frac{1}{2} = 1.5 \]
Since \( f(0) = 1.5 > 0 \), we conclude that \( f(0) \) is positive.

### Step 3: Evaluate \( f\left(\frac{\pi}{2}\right) \)
Calculate \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) - \frac{\pi}{2} + \frac{1}{2} = 0 - \frac{\pi}{2} + \frac{1}{2} \]
Using \( \pi \approx 3.14 \):
\[ f\left(\frac{\pi}{2}\right) \approx -1.57 + 0.5 = -1.07 \]
Since \( f\left(\frac{\pi}{2}\right) < 0 \), we conclude that \( f\left(\frac{\pi}{2}\right) \) is negative.

### Step 4: Apply the Intermediate Value Theorem
Since \( f(0) > 0 \) and \( f\left(\frac{\pi}{2}\right) < 0 \), by the Intermediate Value Theorem, there exists at least one root in the interval \( (0, \frac{\pi}{2}) \).

### Step 5: Check other intervals
1. **For interval B: \( [-\frac{\pi}{2}, 0] \)**:
- Evaluate \( f\left(-\frac{\pi}{2}\right) \):
\[ f\left(-\frac{\pi}{2}\right) = \cos\left(-\frac{\pi}{2}\right) + \frac{\pi}{2} + \frac{1}{2} = 0 + \frac{\pi}{2} + \frac{1}{2} \]
Since \( \frac{\pi}{2} + \frac{1}{2} > 0 \), \( f\left(-\frac{\pi}{2}\right) > 0 \).

- Evaluate \( f(0) \):
\[ f(0) = 1.5 > 0 \]
Thus, both endpoints are positive, indicating no root in this interval.

2. **For interval C: \( [\frac{\pi}{2}, 0] \)**:
- As previously calculated, \( f\left(\frac{\pi}{2}\right) < 0 \) and \( f(0) > 0 \). This does not provide a valid interval since it is not properly defined.

### Conclusion
The only interval where we found a change in sign (indicating a root) is in option A: \( [0, \frac{\pi}{2}] \).

Thus, the answer is:
**(A) \([0, \frac{\pi}{2}]\)**.