Solutions of Trigonometric Equation in Open Interval

To determine the number of solutions to the equation

$$\sin x + 2 \sin 2x - \sin 3x = 3$$

for $x \in (0, \pi)$, we can analyze the maximum possible value of the left-hand side.

Let

$$f(x) = \sin x + 2 \sin 2x - \sin 3x.$$

Using the trigonometric identity $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$, we rewrite the first and third terms:

$$\sin x - \sin 3x = -2 \sin x \cos 2x.$$

Substituting this back into $f(x)$:

$$f(x) = 2 \sin 2x - 2 \sin x \cos 2x.$$

Using $\sin 2x = 2 \sin x \cos x$, we factor out $2 \sin x$:

$$f(x) = 2 \sin x (2 \cos x - \cos 2x).$$

Expressing $\cos 2x$ in terms of $\cos x$ via $\cos 2x = 2 \cos^2 x - 1$:

$$f(x) = 2 \sin x \left(2 \cos x - (2 \cos^2 x - 1)\right) = 2 \sin x (1 + 2 \cos x - 2 \cos^2 x).$$

Now, complete the square for the quadratic term in $\cos x$:

$$1 + 2 \cos x - 2 \cos^2 x = \frac{3}{2} - 2\left(\cos x - \frac{1}{2}\right)^2 \le \frac{3}{2}.$$

Since for all $x \in (0, \pi)$, $\sin x > 0$, we have:

$$f(x) \le 2 \sin x \cdot \frac{3}{2} = 3 \sin x \le 3.$$

For $f(x)$ to attain the value $3$, both of the following conditions must hold simultaneously:

1. $\cos x - \frac{1}{2} = 0 \implies \cos x = \frac{1}{2}$,
2. $\sin x = 1$.

However, if $\cos x = \frac{1}{2}$, then:

$$\sin x = \sqrt{1 - \cos^2 x} = \frac{\sqrt{3}}{2} \neq 1.$$

Thus, both equality conditions cannot be satisfied simultaneously, which implies:

$$f(x) < 3 \quad \text{for all } x \in (0, \pi).$$

Therefore, the equation $\sin x + 2 \sin 2x - \sin 3x = 3$ has no solution in the interval $(0, \pi)$.

Correct Option: (D)