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

The correct option is D no solution

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

We know that,
$\text{LHS}=-(2\cos x-1{)}^2\le 0$
$\text{RHS}=\frac{3}{\sin x}-3\ge 0$

They will be equal when,
$2\cos x-1=0\Rightarrow x=\frac{\pi }{3}$
$\frac{3}{\sin x}-3=0\Rightarrow x=\frac{\pi }{2}$

As the both give different value of $x$,
Therefore, the given equation has no solution.