The number of solutions of the equation `xe^(sinx)-cosx=0` in the interval `(0,pi//2)`, is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To find the number of solutions of the equation \( xe^{\sin x} - \cos x = 0 \) in the interval \( (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Define the function
Let \( f(x) = xe^{\sin x} - \cos x \).

### Step 2: Differentiate the function
We need to find the derivative \( f'(x) \) to analyze the behavior of the function:
\[ f'(x) = \frac{d}{dx}(xe^{\sin x}) - \frac{d}{dx}(\cos x) \]
Using the product rule on \( xe^{\sin x} \):
\[ f'(x) = e^{\sin x} + xe^{\sin x} \cos x + \sin x \]
Thus, we have:
\[ f'(x) = e^{\sin x}(1 + x \cos x) + \sin x \]

### Step 3: Analyze the sign of the derivative
In the interval \( (0, \frac{\pi}{2}) \):
- \( e^{\sin x} > 0 \) since the exponential function is always positive.
- \( \cos x \) is positive in this interval.
- Therefore, \( 1 + x \cos x > 0 \) for \( x > 0 \) (as both terms are positive).
- \( \sin x > 0 \) in this interval.

Thus, \( f'(x) > 0 \) for \( x \in (0, \frac{\pi}{2}) \).

### Step 4: Evaluate the function at the endpoints
Now we evaluate \( f(x) \) at the endpoints of the interval:
- At \( x = 0 \):
\[ f(0) = 0 \cdot e^{\sin 0} - \cos 0 = 0 - 1 = -1 \]
- At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} e^{\sin\left(\frac{\pi}{2}\right)} - \cos\left(\frac{\pi}{2}\right) = \frac{\pi}{2} e^{1} - 0 = \frac{\pi}{2} e \]
Since \( e > 1 \), \( \frac{\pi}{2} e > 0 \).

### Step 5: Conclusion
Since \( f(0) = -1 < 0 \) and \( f\left(\frac{\pi}{2}\right) > 0 \), and since \( f(x) \) is continuous and strictly increasing in the interval \( (0, \frac{\pi}{2}) \), by the Intermediate Value Theorem, there must be exactly one root in the interval \( (0, \frac{\pi}{2}) \).

Thus, the number of solutions of the equation \( xe^{\sin x} - \cos x = 0 \) in the interval \( (0, \frac{\pi}{2}) \) is **1**.