For `x in (0,pi)` the equation `sinx+2sin2x-sin3x=3` has

Options

A. infinitely many solutions
B. three solutions
C. one solution
D. no solution

Explanation

To solve the equation \( \sin x + 2 \sin 2x - \sin 3x = 3 \) for \( x \in (0, \pi) \), we will analyze the left-hand side of the equation and compare it with the right-hand side.

### Step-by-Step Solution:

1. **Understanding the Range of the Functions**:
- The sine function, \( \sin x \), has a maximum value of 1 for \( x \in (0, \pi) \).
- Therefore, \( \sin x \leq 1 \).

2. **Analyzing \( \sin 2x \)**:
- The function \( \sin 2x \) also has a maximum value of 1.
- Thus, \( 2 \sin 2x \leq 2 \).

3. **Analyzing \( \sin 3x \)**:
- Similarly, \( \sin 3x \) has a maximum value of 1.
- Therefore, \( -\sin 3x \geq -1 \).

4. **Combining the Results**:
- Now, we can combine these results to find the maximum value of the left-hand side:
\[ \sin x + 2 \sin 2x - \sin 3x \leq 1 + 2 - (-1) = 1 + 2 + 1 = 4. \]
- However, we need to be more precise. The maximum value of \( \sin x + 2 \sin 2x - \sin 3x \) must be calculated more accurately.

5. **Finding the Maximum Value**:
- The maximum value of \( \sin x + 2 \sin 2x - \sin 3x \) occurs when each term is maximized.
- However, since \( \sin x \) and \( \sin 3x \) cannot both be at their maximum at the same time within the interval \( (0, \pi) \), we need to check specific values of \( x \).

6. **Checking Values**:
- Let's check \( x = \frac{\pi}{2} \):
\[ \sin\left(\frac{\pi}{2}\right) + 2 \sin\left(\pi\right) - \sin\left(\frac{3\pi}{2}\right) = 1 + 0 - (-1) = 2. \]
- This is less than 3.

7. **Conclusion**:
- Since the maximum value of \( \sin x + 2 \sin 2x - \sin 3x \) cannot reach 3, we conclude that the equation \( \sin x + 2 \sin 2x - \sin 3x = 3 \) has no solutions in the interval \( (0, \pi) \).

### Final Answer:
The equation \( \sin x + 2 \sin 2x - \sin 3x = 3 \) has **no solutions** for \( x \in (0, \pi) \).