Question

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

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

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

Using $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

And $\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$ we get

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

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

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

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

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

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

$\Rightarrow \cos (\frac{\pi }{4}+2x)=0$

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

$\Rightarrow x=n\pi \pm \frac{\pi }{4}-\frac{\pi }{8}$ ...(1)

Or $2\cos x+3=0\Rightarrow \cos x=-\frac{3}{2}$

No solution as $-1\le \cos x\le 1$

$\therefore$ For $0\le x\le 2\pi$

$x=\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},\frac{13\pi }{8}$