Question

For $x\in$(0,$\pi$),the equation $sinx+2sin2x-sin3x=3$ has

• A. Infinitely many solution
• B. Three solution
• C. One solution
• D. no solution

Answer 1

Answer: (D)

no solution

solution:

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

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

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

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

$\Rightarrow 0={\cos }^{2}2x+{\sin }^{2}2x+1-2\sin 2x$

$+{\cos }^{2}x+{\sin }^{2}x-2\cos 2x\sin x$

$\Rightarrow 0+{\sin }^{2}2x+1-2\sin 2x+{\cos }^{2}2x+{\sin }^{2}x$

$\Rightarrow 2\cos 2x\sin x+{\cos }^{2}x$

$\Rightarrow 0=(\sin 2x-1{)}^2+(\cos 2x-\sin x{)}^2+{\cos }^{2}x$

Sum of all three non-negative ralues is zero so each term must be zero

$\to \sin 2x-1=0\Rightarrow \sin 2x=1$

$\therefore 2x=\frac{\pi }{2}\Rightarrow x=\frac{\pi }{4}$

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

$\Rightarrow 1-2{\sin }^{2}x-\sin x=0$

$\Rightarrow -2{\sin }^{2}x-2\sin x+1+\sin x=0$

$\Rightarrow -2\sin x(\sin x+1)+1(\sin x+1)=0$

$\to \sin x=-1$ ar $\sin x=+\frac{1}{2}$

$\because x\in (0,\pi )$

where $\sin x>0x=\frac{\pi }{6}$ or $\frac{5\pi }{6}$

always so no solution from this case

$\to {\cos }^{2}x=0\Rightarrow x=\frac{\pi }{2}$

obtained values of $x=\frac{\pi }{4},\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6}$

but we did not get any nalue of x common in each of three cases so each term of equation (1) will never be zero at one particular x

so we got no solution.

Ansmer: option (D)