If `pilexle2pi`, then `cos^(-1)(cosx)` is equal to

Options

A. `x`
B. `pi-x`
C. `2pi+x`
D. `2pi-x`

Explanation

To solve the problem, we need to find the value of \( \cos^{-1}(\cos x) \) given that \( x \) belongs to the interval \( [\pi, 2\pi] \).

### Step-by-Step Solution:

1. **Understanding the Interval**:
We know that \( x \) is in the interval \( [\pi, 2\pi] \). This means that \( x \) can take values from \( \pi \) to \( 2\pi \).

2. **Rewriting \( x \)**:
We can express \( x \) as \( x = \pi + \theta \), where \( \theta \) is in the interval \( [0, \pi] \). This substitution helps us analyze the cosine function more easily.

3. **Finding \( \cos x \)**:
Using the identity for cosine, we have:
\[ \cos x = \cos(\pi + \theta) = -\cos(\theta) \]
This is because \( \cos(\pi + \theta) \) is equal to \(-\cos(\theta)\).

4. **Applying the Inverse Cosine**:
Now we apply the inverse cosine function:
\[ \cos^{-1}(\cos x) = \cos^{-1}(-\cos(\theta)) \]

5. **Using the Inverse Cosine Identity**:
The identity for the inverse cosine states that:
\[ \cos^{-1}(-\cos(\theta)) = \pi - \theta \]
This is valid because \( \theta \) is in the range \( [0, \pi] \).

6. **Substituting Back for \( \theta \)**:
Recall that \( \theta = x - \pi \). Therefore, we can substitute back:
\[ \cos^{-1}(\cos x) = \pi - (x - \pi) = \pi - x + \pi = 2\pi - x \]

7. **Final Result**:
Thus, the final result is:
\[ \cos^{-1}(\cos x) = 2\pi - x \]

### Conclusion:
The value of \( \cos^{-1}(\cos x) \) when \( x \) is in the interval \( [\pi, 2\pi] \) is \( 2\pi - x \).