Question

The equation $\sin x+2\sin 2x=3+\sin 3x$ has no solution in $(0,\pi )$.

• A. True
• B. False

Answer 1

The correct option is A True

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

$-2\sin x+2\sin 2x+4{\sin }^{3}x=3$

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

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

$2\sin x(2\cos x-\cos 2x)=3$

The maximum value attained by $(2\cos x-2\cos 2x)$ is $\frac{3}{2}$ at $x=\frac{\pi }{3}$

At $x=\frac{\pi }{3}$ is $\frac{3\sqrt{3}}{2}<3$. Hence, it is not possible.