One root of the equation `"cos" theta-theta + (1)/(2) = 0` lies in the interval

Options

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

Answer

To solve the equation \( \cos \theta - \theta + \frac{1}{2} = 0 \) and determine the interval in which one of its roots lies, we can follow these steps:

### Step 1: Define the function
We start by defining the function based on the given equation:
\[ f(\theta) = \cos \theta - \theta + \frac{1}{2} \]

### Step 2: Evaluate the function at the endpoints of the interval
We will evaluate \( f(\theta) \) at \( \theta = 0 \) and \( \theta = \frac{\pi}{2} \).

#### Step 2.1: Calculate \( f(0) \)
\[ f(0) = \cos(0) - 0 + \frac{1}{2} = 1 - 0 + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2} \]
Since \( \frac{3}{2} > 0 \), we have:
\[ f(0) > 0 \]

#### Step 2.2: Calculate \( f\left(\frac{\pi}{2}\right) \)
\[ f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) - \frac{\pi}{2} + \frac{1}{2} = 0 - \frac{\pi}{2} + \frac{1}{2} = -\frac{\pi}{2} + \frac{1}{2} \]
Since \( \pi \approx 3.14 \), we can approximate:
\[ -\frac{\pi}{2} + \frac{1}{2} < 0 \]
Thus, we have:
\[ f\left(\frac{\pi}{2}\right) < 0 \]

### Step 3: Apply the Intermediate Value Theorem
Since \( f(0) > 0 \) and \( f\left(\frac{\pi}{2}\right) < 0 \), by the Intermediate Value Theorem, there must be at least one root in the interval \( (0, \frac{\pi}{2}) \).

### Conclusion
Therefore, we conclude that one root of the equation \( \cos \theta - \theta + \frac{1}{2} = 0 \) lies in the interval:
\[ (0, \frac{\pi}{2}) \]