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 where \( x \) is given as \( x \in [-2\pi, 2\pi] \) and we need to find \( \cos^{-1}(\cos x) \), we can follow these steps:

### Step 1: Understanding the Function
The function \( \cos^{-1}(\cos x) \) returns the angle whose cosine is \( \cos x \). However, the output of \( \cos^{-1} \) is restricted to the range \( [0, \pi] \).

### Step 2: Determine the Value of \( x \)
Given that \( x \) is in the interval \( [-2\pi, 2\pi] \), we need to find the equivalent angle of \( x \) that lies within the principal range of the \( \cos^{-1} \) function.

### Step 3: Adjust \( x \) to the Principal Range
To adjust \( x \) to the range of \( [0, \pi] \), we can use the periodic property of the cosine function. Specifically, we can express \( x \) in terms of an equivalent angle:
- If \( x \) is in the interval \( [0, \pi] \), then \( \cos^{-1}(\cos x) = x \).
- If \( x \) is in the interval \( [\pi, 2\pi] \), then \( \cos^{-1}(\cos x) = 2\pi - x \).
- If \( x \) is in the interval \( [-\pi, 0] \), then \( \cos^{-1}(\cos x) = -x \).

### Step 4: Apply the Range
Since \( x \) can be negative or positive, we will consider the cases:
- For \( x \in [-2\pi, -\pi] \): \( \cos^{-1}(\cos x) = -x \).
- For \( x \in [-\pi, 0] \): \( \cos^{-1}(\cos x) = -x \).
- For \( x \in [0, \pi] \): \( \cos^{-1}(\cos x) = x \).
- For \( x \in [\pi, 2\pi] \): \( \cos^{-1}(\cos x) = 2\pi - x \).

### Step 5: Conclusion
Thus, for \( x \in [-2\pi, 2\pi] \):
- If \( x \) is in the interval \( [-2\pi, -\pi] \), \( \cos^{-1}(\cos x) = -x \).
- If \( x \) is in the interval \( [-\pi, 0] \), \( \cos^{-1}(\cos x) = -x \).
- If \( x \) is in the interval \( [0, \pi] \), \( \cos^{-1}(\cos x) = x \).
- If \( x \) is in the interval \( [\pi, 2\pi] \), \( \cos^{-1}(\cos x) = 2\pi - x \).

### Final Answer
Therefore, the answer to the question is:
\[ \cos^{-1}(\cos x) = \begin{cases} -x & \text{if } x \in [-2\pi, -\pi] \\ -x & \text{if } x \in [-\pi, 0] \\ x & \text{if } x \in [0, \pi] \\ 2\pi - x & \text{if } x \in [\pi, 2\pi] \end{cases} \]