The number of solutions of `2"cos"^(2)((x)/(2))"sin"^(2) x =x^(2) + (1)/(x^(2)), 0 le x le (pi)/(2),` is

Options

A. 0
B. 1
C. infinite
D. none of these

Answer

To solve the equation \( 2 \cos^2\left(\frac{x}{2}\right) \sin^2 x = x^2 + \frac{1}{x^2} \) for \( 0 \leq x \leq \frac{\pi}{2} \), we will analyze both sides of the equation step by step.

### Step 1: Analyze the right-hand side
We start with the right-hand side of the equation, \( x^2 + \frac{1}{x^2} \).

Using the AM-GM inequality:
\[ x^2 + \frac{1}{x^2} \geq 2 \]
This inequality holds for all \( x > 0 \). The equality holds when \( x^2 = 1 \) or \( x = 1 \).

### Step 2: Analyze the left-hand side
Now, we analyze the left-hand side, \( 2 \cos^2\left(\frac{x}{2}\right) \sin^2 x \).

The maximum values of \( \cos^2\left(\frac{x}{2}\right) \) and \( \sin^2 x \) are both 1. Therefore, the maximum value of the left-hand side is:
\[ 2 \cdot 1 \cdot 1 = 2 \]

### Step 3: Set up the inequality
From the previous steps, we have:
\[ 2 \cos^2\left(\frac{x}{2}\right) \sin^2 x \leq 2 \]
and
\[ x^2 + \frac{1}{x^2} \geq 2 \]

### Step 4: Equate both sides
For the equality \( 2 \cos^2\left(\frac{x}{2}\right) \sin^2 x = x^2 + \frac{1}{x^2} \) to hold, both sides must equal 2.

1. For \( 2 \cos^2\left(\frac{x}{2}\right) \sin^2 x = 2 \):
- This occurs when \( \cos^2\left(\frac{x}{2}\right) = 1 \) and \( \sin^2 x = 1 \).
- \( \cos^2\left(\frac{x}{2}\right) = 1 \) implies \( \frac{x}{2} = n\pi \) for integer \( n \), which gives \( x = 0 \) (since \( 0 \leq x \leq \frac{\pi}{2} \)).
- \( \sin^2 x = 1 \) implies \( x = \frac{\pi}{2} \).

2. Check these values:
- For \( x = 0 \): \( 2 \cos^2(0) \sin^2(0) = 0 \) and \( 0^2 + \frac{1}{0^2} \) is undefined.
- For \( x = \frac{\pi}{2} \): \( 2 \cos^2\left(\frac{\pi}{4}\right) \sin^2\left(\frac{\pi}{2}\right) = 2 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 \cdot 1 = 1 \) and \( \left(\frac{\pi}{2}\right)^2 + \frac{1}{\left(\frac{\pi}{2}\right)^2} > 2 \).

### Conclusion
Since neither \( x = 0 \) nor \( x = \frac{\pi}{2} \) satisfies the equation, we conclude that there are no solutions in the interval \( 0 \leq x \leq \frac{\pi}{2} \).

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