For $x \in (0, \pi)$, the equation $\sin x + 2\sin 2x - \sin 3x = 3$ has

For x ∈ (0, π), the equation sin x + 2sin 2x - sin 3x = 3 has

$\text{Given: } x \in (0,\pi), \quad \sin x + 2\sin 2x - \sin 3x = 3$

$\text{To find: number of solutions for } x \in (0,\pi)$

$\text{Options: infinite},\ 3,\ 1,\ \text{none}$

$\sin 2x = 2\sin x\cos x$

$\sin 3x = \sin x\left(4\cos^{2}x-1\right)$

$\text{LHS} = \sin x + 4\sin x\cos x - \sin x\left(4\cos^{2}x-1\right)$

$= \sin x\left(1 + 4\cos x - 4\cos^{2}x + 1\right)$

$= \sin x\left(2 + 4\cos x - 4\cos^{2}x\right)$

$2 + 4\cos x - 4\cos^{2}x = 3 - 4\left(\cos x-\frac{1}{2}\right)^{2}$

$\text{LHS} = 3\sin x - 4\sin x\left(\cos x-\frac{1}{2}\right)^{2}$

$\sin x > 0, \quad \left(\cos x-\frac{1}{2}\right)^{2} \geq 0$

$\text{LHS} \leq 3\sin x \leq 3$

$\text{Equality would need } \sin x = 1 \ \text{and} \ \cos x = \frac{1}{2}$

$\text{impossible} \Rightarrow \text{LHS} < 3$

$x = \frac{\pi}{2}: \ \text{LHS} = 1 + 0 + 1 = 2 < 3$

$\sin x + 2\sin 2x - \sin 3x = 3 \Rightarrow \text{no solution}$

$\text{(4) no solution}$