For `x in (0, pi)`, the equation
`"sin"x + 2"sin" 2x-"sin" 3x = 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 the interval \( (0, \pi) \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin x + 2 \sin 2x - \sin 3x = 3 \]

### Step 2: Use trigonometric identities
Recall the double angle and triple angle identities:
- \( \sin 2x = 2 \sin x \cos x \)
- \( \sin 3x = 3 \sin x - 4 \sin^3 x \)

Substituting these identities into the equation gives:
\[ \sin x + 2(2 \sin x \cos x) - (3 \sin x - 4 \sin^3 x) = 3 \]
This simplifies to:
\[ \sin x + 4 \sin x \cos x - 3 \sin x + 4 \sin^3 x = 3 \]

### Step 3: Combine like terms
Now, combine the terms involving \( \sin x \):
\[ 4 \sin x \cos x + 4 \sin^3 x - 2 \sin x = 3 \]
Rearranging gives:
\[ 4 \sin x \cos x + 4 \sin^3 x - 3 = 2 \sin x \]

### Step 4: Move all terms to one side
Rearranging the equation leads to:
\[ 4 \sin^3 x + 4 \sin x \cos x - 2 \sin x - 3 = 0 \]

### Step 5: Factor out \( \sin x \)
Factoring out \( \sin x \) from the first two terms gives:
\[ \sin x (4 \sin^2 x + 4 \cos x - 2) - 3 = 0 \]

### Step 6: Analyze the equation
This equation can be split into two parts:
1. \( \sin x = 0 \)
2. \( 4 \sin^2 x + 4 \cos x - 2 = 3 \)

### Step 7: Solve \( \sin x = 0 \)
In the interval \( (0, \pi) \), \( \sin x = 0 \) has no solutions since \( x = 0 \) and \( x = \pi \) are not included in the open interval.

### Step 8: Solve the second equation
Now, we need to solve:
\[ 4 \sin^2 x + 4 \cos x - 2 = 3 \]
This simplifies to:
\[ 4 \sin^2 x + 4 \cos x - 5 = 0 \]

### Step 9: Substitute \( \sin^2 x \) and \( \cos^2 x \)
Using \( \sin^2 x + \cos^2 x = 1 \), we can express \( \sin^2 x \) in terms of \( \cos x \):
\[ \sin^2 x = 1 - \cos^2 x \]
Substituting this into the equation gives:
\[ 4(1 - \cos^2 x) + 4 \cos x - 5 = 0 \]
This simplifies to:
\[ -4 \cos^2 x + 4 \cos x - 1 = 0 \]

### Step 10: Solve the quadratic equation
Rearranging gives:
\[ 4 \cos^2 x - 4 \cos x + 1 = 0 \]
Using the quadratic formula:
\[ \cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \]
Calculating the discriminant:
\[ 16 - 16 = 0 \]
Thus, there is one solution:
\[ \cos x = \frac{4}{8} = \frac{1}{2} \]

### Step 11: Find \( x \)
The solution \( \cos x = \frac{1}{2} \) gives:
\[ x = \frac{\pi}{3} \]

### Conclusion
The only solution in the interval \( (0, \pi) \) is:
\[ x = \frac{\pi}{3} \]