The number of real solutions of the equation
`sqrt(1+cos2x) = sqrt(2)cos^(-1)(cosx)` in `[pi/2,pi]` is

Options

A. 0
B. 1
C. 2
D. `oo`

Explanation

To solve the equation \( \sqrt{1 + \cos 2x} = \sqrt{2} \cos^{-1}(\cos x) \) in the interval \([ \frac{\pi}{2}, \pi ]\), we will follow these steps:

### Step 1: Simplify the left-hand side
We know that \( \cos 2x = 2 \cos^2 x - 1 \). Therefore, we can rewrite \( 1 + \cos 2x \) as follows:
\[ 1 + \cos 2x = 1 + (2 \cos^2 x - 1) = 2 \cos^2 x \]
Thus, we have:
\[ \sqrt{1 + \cos 2x} = \sqrt{2 \cos^2 x} = \sqrt{2} \cos x \]

### Step 2: Simplify the right-hand side
The right-hand side is \( \sqrt{2} \cos^{-1}(\cos x) \). The function \( \cos^{-1}(\cos x) \) gives us \( x \) in the range where \( x \) is in \([0, \pi]\). However, since we are considering \( x \) in the interval \([ \frac{\pi}{2}, \pi ]\), we can write:
\[ \cos^{-1}(\cos x) = x \quad \text{for } x \in [\frac{\pi}{2}, \pi] \]
Thus, the right-hand side simplifies to:
\[ \sqrt{2} \cos^{-1}(\cos x) = \sqrt{2} x \]

### Step 3: Set the simplified expressions equal
Now we have:
\[ \sqrt{2} \cos x = \sqrt{2} x \]
Dividing both sides by \( \sqrt{2} \) (since \( \sqrt{2} \neq 0 \)):
\[ \cos x = x \]

### Step 4: Analyze the equation \( \cos x = x \)
We need to find the number of solutions to the equation \( \cos x = x \) in the interval \([ \frac{\pi}{2}, \pi ]\).

1. At \( x = \frac{\pi}{2} \):
\[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad x = \frac{\pi}{2} \approx 1.57 \]
Here, \( \cos\left(\frac{\pi}{2}\right) < \frac{\pi}{2} \).

2. At \( x = \pi \):
\[ \cos(\pi) = -1 \quad \text{and} \quad x = \pi \approx 3.14 \]
Here, \( \cos(\pi) < \pi \).

3. The function \( \cos x \) is decreasing in the interval \([ \frac{\pi}{2}, \pi ]\) and starts from 0 at \( x = \frac{\pi}{2} \) and goes to -1 at \( x = \pi \). The line \( y = x \) is increasing and starts from \( \frac{\pi}{2} \) and goes to \( \pi \).

### Step 5: Conclusion
Since \( \cos x \) starts at 0 and ends at -1 while \( x \) starts at \( \frac{\pi}{2} \) and goes to \( \pi \), they do not intersect in the interval \([ \frac{\pi}{2}, \pi ]\). Therefore, there are no solutions.

Thus, the number of real solutions of the equation is **0**.