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)

no solution

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

$\sin x+4\sin x\cos x-3\sin x+4{\sin }^{3}x=3$

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

$\sin x[2-(4{\cos }^{2}x-4\cos x+1)+1]=3$

$\sin x[3-(2\cos x-1{)}^2]=3$

$\Rightarrow \sin x=1$ and $2\cos x-1=0$

$\Rightarrow x=\frac{\pi }{2}$ and $x=\frac{\pi }{3}$

Which is not possible at same time.

Hence, no solution.