` x_(1) and x_(2)` are two solutions of the equation ` e^(x) cos x=1 ` , The minimum number of the solution of the equation ` e^(x) sin x =1`, lying between ` x_(1) and x_(2)` can be

Options

A. 0
B. 1
C. 3
D. None of these

Answer

To solve the problem, we need to analyze the given equations and apply some mathematical principles.

### Step-by-Step Solution:

1. **Identify the Functions**:
We have two equations:
- \( e^x \cos x = 1 \)
- \( e^x \sin x = 1 \)

Let’s denote:
- \( f(x) = e^x \cos x - 1 \)
- \( g(x) = e^x \sin x - 1 \)

2. **Find the Solutions for \( f(x) = 0 \)**:
We know that \( x_1 \) and \( x_2 \) are the solutions to the equation \( f(x) = 0 \). According to the problem, these solutions exist in the interval \( (x_1, x_2) \).

3. **Apply Rolle's Theorem**:
Since \( f(x_1) = f(x_2) = 0 \) and \( f(x) \) is continuous and differentiable, by Rolle's theorem, there exists at least one \( c \) in the interval \( (x_1, x_2) \) such that \( f'(c) = 0 \).

4. **Differentiate \( f(x) \)**:
We compute the derivative:
\[ f'(x) = \frac{d}{dx}(e^x \cos x) = e^x \cos x - e^x \sin x = e^x (\cos x - \sin x) \]
Setting this equal to zero gives:
\[ e^x (\cos x - \sin x) = 0 \]
Since \( e^x \) is never zero, we have:
\[ \cos x - \sin x = 0 \implies \tan x = 1 \implies x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

5. **Find the Critical Points**:
The critical points where \( f'(x) = 0 \) are where \( x = \frac{\pi}{4} + n\pi \).

6. **Analyze \( g(x) = 0 \)**:
Now we need to find the solutions for \( g(x) = 0 \):
\[ e^x \sin x = 1 \]
This function will also have critical points, and we can analyze the behavior of \( g(x) \) in the interval \( (x_1, x_2) \).

7. **Minimum Solutions Between \( x_1 \) and \( x_2 \)**:
Since \( f(x) \) has two solutions \( x_1 \) and \( x_2 \), and we found that \( f'(x) = 0 \) at points where \( \tan x = 1 \), we can analyze how many times \( g(x) = 0 \) can cross the x-axis between \( x_1 \) and \( x_2 \).

The function \( g(x) \) will have at least one solution in the interval \( (x_1, x_2) \) because it is continuous, and it will cross the x-axis at least once due to the behavior of sine and the exponential function.

8. **Conclusion**:
Since \( g(x) \) can only cross the x-axis once between \( x_1 \) and \( x_2 \), the minimum number of solutions of the equation \( e^x \sin x = 1 \) lying between \( x_1 \) and \( x_2 \) is **1**.

### Final Answer:
The minimum number of solutions of the equation \( e^x \sin x = 1 \) lying between \( x_1 \) and \( x_2 \) is **1**.