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

Answer

To solve the equation \( \sin x + 2 \sin 2x - \sin 3x = 3 \) for \( x \in (0, \pi) \), we will follow these steps:

### Step 1: Substitute the trigonometric identities
Recall the identities:
- \( \sin 2x = 2 \sin x \cos x \)
- \( \sin 3x = 3 \sin x - 4 \sin^3 x \)

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

### Step 2: Simplify the equation
Now, simplify the equation:
\[ \sin x + 4 \sin x \cos x - 3 \sin x + 4 \sin^3 x = 3 \]
This simplifies to:
\[ (4 \sin x \cos x + 4 \sin^3 x - 2 \sin x) = 3 \]

### Step 3: Factor out \( \sin x \)
Factor out \( \sin x \):
\[ \sin x (4 \cos x + 4 \sin^2 x - 2) = 3 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 4 \sin x \cos x + 4 \sin^3 x - 2 \sin x - 3 = 0 \]

### Step 5: Analyze the equation
To find the solutions, we need to analyze the left-hand side. The left-hand side must equal 3, which is the maximum value for \( \sin x \) in the interval \( (0, \pi) \).

### Step 6: Determine the maximum values
The maximum value of \( \sin x \) is 1, which occurs at \( x = \frac{\pi}{2} \). At this point:
\[ \sin x = 1 \implies 4 \cdot 1 \cdot \cos\left(\frac{\pi}{2}\right) + 4 \cdot 1^3 - 2 \cdot 1 = 4 - 2 = 2 \]
This is less than 3.

### Step 7: Check other possible values
Next, check \( \sin x \) values at \( x = \frac{\pi}{3} \):
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
Substituting these values:
\[ 4 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} + 4 \cdot \left(\frac{\sqrt{3}}{2}\right)^3 - 2 \cdot \frac{\sqrt{3}}{2} \]
Calculating this gives:
\[ 4 \cdot \frac{\sqrt{3}}{4} + 4 \cdot \frac{3\sqrt{3}}{8} - \sqrt{3} = \sqrt{3} + \frac{3\sqrt{3}}{2} - \sqrt{3} = \frac{3\sqrt{3}}{2} \]
This is also less than 3.

### Step 8: Conclusion
Since the left-hand side cannot reach 3 for any \( x \in (0, \pi) \), we conclude that the equation has no solutions in the given interval.

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

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