Question

The solutions of the equation $\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$ in interval $0\le x\le 2\pi$ are :

• A. $\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8}$
• B. $\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},\frac{13\pi }{8}$
• C. $\frac{4\pi }{3},\frac{9\pi }{3},\frac{2\pi }{3},\frac{13\pi }{8}$
• D. $\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{3},\frac{4\pi }{3}$

Answer 1

The correct option is B $\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},\frac{13\pi }{8}$

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

$\Rightarrow (\sin x+\sin 3x)-3\sin 2x-(\cos x+\cos 3x)+3\cos 2x=0$

$\Rightarrow 2\sin \left(\frac{x+3x}{2}\right)\cos \left(\frac{x-3x}{2}\right)-3\sin 2x-2\cos \left(\frac{x+3x}{2}\right)\cos \left(\frac{x-3x}{2}\right)+3\cos 2x=0$

$\Rightarrow 2\sin 2x\cos x-3\sin 2x-2\cos 2x\cos x+3\cos 2x=0$

$\Rightarrow \sin 2x(2\cos x-3)-\cos 2x(2\cos x-3)=0$

$\Rightarrow (2\cos x-3)(\sin 2x-\cos 2x)=0$

$\Rightarrow \cos x=\frac{3}{2}$ or $\sin 2x=\cos 2x$

As $\cos x\in [-1,1]$

Thus,$\cos x\ne \frac{3}{2}$

So, $\sin 2x=\cos 2x$

$\Rightarrow \tan 2x=1$

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

$\Rightarrow 2x=n\pi +\frac{\pi }{4}$

$\Rightarrow x=\frac{n\pi }{2}+\frac{\pi }{8}$

For $n=0,x=\frac{\pi }{8}$

For $n=1,x=\frac{\pi }{2}+\frac{\pi }{8}=\frac{4\pi +\pi }{8}=\frac{5\pi }{8}$

For $n=2,x=\pi +\frac{\pi }{8}=\frac{9\pi }{8}$

For $n=3,x=\frac{3\pi }{2}+\frac{\pi }{8}=\frac{12\pi +\pi }{8}=\frac{13\pi }{8}$

Thus, the solutions are $\{\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},\frac{13\pi }{8}\}$