Question

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

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

Answer 1

The correct option is D no solution

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

$\sin x+4\sin x\cos x=3+\sin 3x$

$\sin x(1+4\cos x)=3+\sin 3x$

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

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

$4{\cos }^{2}x-4\cos x-2=9$

$(2\cos x-1{)}^2-3\ge -3$

and on $(0,\pi )$ we have $0<\sin x\le 1$ they

The $LHS$ of your last equation. should be $\ge -3$ on $(0,\pi )$

No Solution in $(0,\pi )$