The number of real solution of the equation `sqrt(1 + cos 2x) = sqrt2 sin^(-1) (sin x), -pi le x le pi`, is

Options

A. 0
B. 1
C. 2
D. infinite

Explanation

To solve the equation \( \sqrt{1 + \cos 2x} = \sqrt{2} \sin^{-1}(\sin x) \) for \( -\pi \leq x \leq \pi \), we will follow these steps:

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

### Step 2: Simplify the right-hand side
The right-hand side is:
\[ \sqrt{2} \sin^{-1}(\sin x) \]
For \( -\pi \leq x \leq \pi \), \( \sin^{-1}(\sin x) = x \) when \( x \) is in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) and \( \sin^{-1}(\sin x) = \pi - x \) when \( x \) is in the range \( [\frac{\pi}{2}, \pi] \) or \( -\pi \leq x < -\frac{\pi}{2} \).

### Step 3: Analyze cases based on the range of \( x \)

#### Case 1: \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \)
In this range, we have:
\[ \sqrt{2} |\cos x| = \sqrt{2} x \]
Dividing both sides by \( \sqrt{2} \) (since \( \sqrt{2} > 0 \)):
\[ |\cos x| = x \]
Since \( \cos x \) is non-negative in this interval, we can write:
\[ \cos x = x \]
To find the number of solutions, we can analyze the graphs of \( y = \cos x \) and \( y = x \). They intersect at one point in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

#### Case 2: \( -\pi \leq x < -\frac{\pi}{2} \)
In this range, we have:
\[ \sqrt{2} (-\cos x) = \sqrt{2} (\pi - x) \]
Dividing both sides by \( \sqrt{2} \):
\[ -\cos x = \pi - x \]
Rearranging gives:
\[ \cos x = x - \pi \]
The function \( \cos x \) is decreasing and \( x - \pi \) is a linear function. We can analyze the intersection of these two functions, which will yield no solutions in this interval since \( \cos x \) is always less than or equal to 1, while \( x - \pi < -1 \).

#### Case 3: \( \frac{\pi}{2} < x \leq \pi \)
In this range, we have:
\[ -\sqrt{2} \cos x = \sqrt{2} (\pi - x) \]
Dividing both sides by \( \sqrt{2} \):
\[ -\cos x = \pi - x \]
Rearranging gives:
\[ \cos x = x - \pi \]
Again, we analyze the intersection of these two functions. The cosine function decreases and the linear function \( x - \pi \) will intersect it once in this interval.

### Step 4: Count the solutions
From our analysis:
- In the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), there is **1 solution**.
- In the interval \( [-\pi, -\frac{\pi}{2}) \), there are **0 solutions**.
- In the interval \( (\frac{\pi}{2}, \pi] \), there is **1 solution**.

### Conclusion
The total number of real solutions to the equation \( \sqrt{1 + \cos 2x} = \sqrt{2} \sin^{-1}(\sin x) \) in the interval \( -\pi \leq x \leq \pi \) is:
\[ \text{Total Solutions} = 1 + 0 + 1 = 2 \]

Thus, the number of real solutions is **2**.