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 \( x \in (0, \pi) \), we will follow these steps:

### Step 1: Rewrite the equation
The equation given is:
\[ \sin x + 2 \sin 2x - \sin 3x = 3 \]
We can use 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:
\[ \sin x + 2(2 \sin x \cos x) - (3 \sin x - 4 \sin^3 x) = 3 \]

### Step 2: Simplify the equation
Now, simplify the equation:
\[ \sin x + 4 \sin x \cos x - 3 \sin x + 4 \sin^3 x = 3 \]
Combine like terms:
\[ -2 \sin x + 4 \sin x \cos x + 4 \sin^3 x = 3 \]

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

### Step 4: Substitute \( \sin^2 x \)
Recall that \( \sin^2 x = 1 - \cos^2 x \). Substitute this into the equation:
\[ \sin x (-2 + 4 \cos x + 4(1 - \cos^2 x)) = 3 \]
This simplifies to:
\[ \sin x (2 + 4 \cos x - 4 \cos^2 x) = 3 \]

### Step 5: Analyze the equation
The left side of the equation, \( \sin x (2 + 4 \cos x - 4 \cos^2 x) \), must equal 3. However, since \( \sin x \) has a maximum value of 1 in the interval \( (0, \pi) \), the maximum value of the left-hand side is:
\[ \sin x \leq 1 \implies \sin x (2 + 4 \cos x - 4 \cos^2 x) \leq 2 + 4 \cos x - 4 \cos^2 x \]
To find the maximum of \( 2 + 4 \cos x - 4 \cos^2 x \), we can analyze it further.

### Step 6: Check the maximum value
The expression \( 2 + 4 \cos x - 4 \cos^2 x \) is a quadratic in \( \cos x \):
\[ -4 \cos^2 x + 4 \cos x + 2 \]
The maximum value of a quadratic \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \):
\[ \cos x = -\frac{4}{2(-4)} = \frac{1}{2} \]
Substituting \( \cos x = \frac{1}{2} \):
\[ 2 + 4 \cdot \frac{1}{2} - 4 \cdot \left(\frac{1}{2}\right)^2 = 2 + 2 - 1 = 3 \]
Thus, the maximum value of the left-hand side can equal 3 when \( \sin x = 1 \) and \( \cos x = \frac{1}{2} \).

### Step 7: Solve for \( x \)
The only \( x \) in the interval \( (0, \pi) \) that satisfies \( \sin x = 1 \) is:
\[ x = \frac{\pi}{2} \]
However, substituting \( x = \frac{\pi}{2} \) back into the original equation gives:
\[ \sin\left(\frac{\pi}{2}\right) + 2 \sin(\pi) - \sin\left(\frac{3\pi}{2}\right) = 1 + 0 + 1 = 2 \neq 3 \]
Thus, there is no solution in the interval \( (0, \pi) \).

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

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