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

$\sin x+3\sin 2x+\sin 3x=\cos x+3\cos 2x+\cos 3x$
$formulaused$
$\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}$
$\cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}$
$\cos (-\theta )=\cos \theta$
$\Rightarrow 2\sin 2x\cos x+3\sin 2x=2\cos 2x\cos x+3\cos 2x$
$\Rightarrow \sin 2x[2\cos x+3]=\cos 2x[2\cos x+3]$
$\Rightarrow \sin 2x=\cos 2x$
$\Rightarrow \tan 2x=1=\tan \frac{\pi }{4}$
$\therefore 2x=n\pi +\frac{\pi }{4}$
$x=\frac{n\pi }{2}+\frac{\pi }{8}$
$0\le x\le 2\pi$
$\therefore x=\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},\frac{13\pi }{8}$