Question

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

• A. no solution
• B. one solution
• C. two solutions
• D. three solutions

Answer 1

The correct option is A no solution

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

Maximum value of $3-(2\cos x-1{)}^2$ is $3$ when $\cos x=\frac{1}{2}$ but at this value, $\sin x\ne 1$
Hence, $\sin x[3-(2\cos x-1{)}^2]=3$ is not possible.