Given equation

We need to solve, for $x \in (0,\pi)$, $\sin x + 2\sin 2x - \sin 3x = 3.$

Simplify the left-hand side

Use the identities: $\sin 2x = 2\sin x\cos x,$ $\sin 3x = 3\sin x - 4\sin^3 x.$

Then

Use the condition $x\in(0,\pi)$

For $x\in(0,\pi)$, $\sin x > 0.$

Let $t = \cos x, \qquad -1 < t < 1.$

Then $\sin x = \sqrt{1-t^2}.$

So the left-hand side is $2\sqrt{1-t^2}(1+2t-2t^2).$

We need to check whether this can equal $3$.

A sharper simplification using sum-to-product

Observe: $\sin x - \sin 3x = 2\cos\left(\frac{x+3x}{2}\right)\sin\left(\frac{x-3x}{2}\right)$ $= 2\cos 2x\sin(-x) = -2\sin x\cos 2x.$

Hence

Find the maximum possible value of the left-hand side

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

Using the identity derived earlier, $f(x)=2\sin x(1+2\cos x-2\cos^2 x).$

Put $t=\cos x$, so $f(x)=2\sqrt{1-t^2}(1+2t-2t^2), \qquad -1<t<1.$

Now test whether $f(x)$ can reach $3$.

Since $0<\sin x\le 1$, we estimate the factor $g(t)=1+2t-2t^2.$

This is a downward-opening quadratic. Its maximum occurs at $t = -\frac{b}{2a} = -\frac{2}{2(-2)}=\frac12.$

Then $g\left(\frac12\right)=1+2\cdot\frac12-2\cdot\frac14=1+1-\frac12=\frac32.$

Therefore, $f(x)=2\sin x\, g(t) \le 2\cdot 1 \cdot \frac32 = 3.$

So the left side is at most $3$.

Check when equality can hold

For $f(x)=2\sin x\,g(t)=3$ to happen, we must have equality in both bounds simultaneously:

• $\sin x = 1 \Rightarrow x=\frac\pi2$,
• $g(t)$ maximum at $t=\frac12 \Rightarrow \cos x=\frac12 \Rightarrow x=\frac\pi3$.

These cannot happen at the same time.

Hence the value $3$ is never attained for any $x\in(0,\pi)$.

Conclusion

Therefore, the equation has no solution in $(0,\pi)$.

So the correct option is: $\boxed{\text{D: no solution}}.$