Question

The of solution of the equation sin x + sin 2x +sin 3x = cos x+ cos 2x +cos 3x in $0\le x\le 2\pi$ is

Answer 1

We have,

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

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

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

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

$\Rightarrow \sin 2x(2\cos x+1)=2\cos 2x(\cos x+1)$

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

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

If

$2\cos x+1=0$

$2\cos x=-1$

$\cos x=-\frac{1}{2}$

$\cos x=-\cos \frac{\pi }{3}$

$\cos x=\cos (\pi -\frac{\pi }{3})$

$x=\frac{2\pi }{3}$

If

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

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

$\Rightarrow \tan 2x=1$

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

$\Rightarrow x=\frac{\pi }{8}$

Hence, this is the answer.