The equation `2cos^(2)(x/2) sin^2x=x^(2)+x^(-2), 0 lt x^(-2), 0 lt x le pi/2` has

Options

A. one real solutions
B. more than one real solutions
C. no real solutions
D. none of the above

Answer

To solve the equation \( 2 \cos^2\left(\frac{x}{2}\right) \sin^2(x) = x^2 + x^{-2} \) for \( 0 < x \leq \frac{\pi}{2} \), we will follow these steps:

### Step 1: Rewrite the Left-Hand Side
We start with the left-hand side of the equation:
\[ 2 \cos^2\left(\frac{x}{2}\right) \sin^2(x) \]
Using the identity \( \sin(x) = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \), we can rewrite \( \sin^2(x) \):
\[ \sin^2(x) = \left(2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)\right)^2 = 4 \sin^2\left(\frac{x}{2}\right) \cos^2\left(\frac{x}{2}\right) \]
Now substituting this back into the left-hand side:
\[ 2 \cos^2\left(\frac{x}{2}\right) \cdot 4 \sin^2\left(\frac{x}{2}\right) \cos^2\left(\frac{x}{2}\right) = 8 \cos^4\left(\frac{x}{2}\right) \sin^2\left(\frac{x}{2}\right) \]

### Step 2: Set the Equation
Now we set the left-hand side equal to the right-hand side:
\[ 8 \cos^4\left(\frac{x}{2}\right) \sin^2\left(\frac{x}{2}\right) = x^2 + x^{-2} \]

### Step 3: Analyze the Right-Hand Side
The right-hand side \( x^2 + x^{-2} \) can be rewritten as:
\[ x^2 + \frac{1}{x^2} = \frac{x^4 + 1}{x^2} \]
This expression is always greater than or equal to 2 for \( x > 0 \) (by AM-GM inequality).

### Step 4: Evaluate the Left-Hand Side
Next, we need to evaluate the left-hand side. Notice that:
- \( \cos^2\left(\frac{x}{2}\right) \) and \( \sin^2\left(\frac{x}{2}\right) \) are both non-negative for \( 0 < x \leq \frac{\pi}{2} \).
- The maximum value of \( 8 \cos^4\left(\frac{x}{2}\right) \sin^2\left(\frac{x}{2}\right) \) occurs when \( x = \frac{\pi}{2} \).

### Step 5: Check the Boundaries
At \( x = \frac{\pi}{2} \):
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus,
\[ 8 \left(\frac{1}{\sqrt{2}}\right)^4 \left(\frac{1}{\sqrt{2}}\right)^2 = 8 \cdot \frac{1}{4} \cdot \frac{1}{2} = 1 \]

### Step 6: Compare Both Sides
Now we compare:
\[ 1 \quad \text{(LHS at } x = \frac{\pi}{2}\text{)} \quad \text{with} \quad x^2 + x^{-2} \quad \text{(which is } \geq 2 \text{ for } x > 0\text{)} \]
Since \( 1 < 2 \), it shows that the left-hand side cannot equal the right-hand side.

### Conclusion
Thus, the equation \( 2 \cos^2\left(\frac{x}{2}\right) \sin^2(x) = x^2 + x^{-2} \) has **no solutions** in the interval \( 0 < x \leq \frac{\pi}{2} \).