Question

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

• A. infinitely many solutions
• B. three solutions
• C. one solution
• D. no solution

Answer 1

Answer: (D)

$\sin x+2\sin 2x-\sin 3x=3$
$\sin x+4\sin x\cos x-3\sin x+4{\sin }^{3}x=3$
$\sin x\left[-2+4\cos x+4\left(1-{\cos }^{2}x\right)\right]=3$
$\sin x\left[2-\left(4{\cos }^{2}x-4\cos x+1\right)+1\right]=3$
$\sin x\left[3-(2\cos x-1{)}^2\right]=3$
$\Rightarrow \sin x=1$ and $2\cos x-1=0$
$\Rightarrow x=\frac{\pi }{2}$ and $x=\frac{\pi }{3}$