Question

Solve for $x:\sin x+\sin 2x+\sin 3x=\cos x+\cos 2x+\cos 3x$ in the interval $0\le x\le 2\pi$.

Answer 1

We write the given equation as
$(\sin x+\sin 3x)+\sin 2x=(\cos x+\cos 3x)+\cos 2x$
or $2\sin 2x\cos x+\sin 2x=2\cos 2x\cos x+\cos 2x$
or $\sin 2x(2\cos x+1)=\cos 2x(2\cos x+1)$
or $(\sin 2x-\cos 2x)(2\cos x+1)=0$
$\therefore \sin 2x-\cos 2x=0$ or $2\cos x+1=0$
If $\sin 2x-\cos 2x=0$, then $\tan 2x=1$,
Hence $2x=n\pi +\pi /4$
or $x=(4n+1)\frac{\pi }{8}$ .$\left(1\right)$
If $2\cos x+1=0$, then $\cos x=-1/2$ $\left(2\right)$
$\therefore x=2n\pi \pm \frac{2\pi }{3}$ or $x=\frac{6n\pm 2}{3}\pi$
We seek values of x in the interval $0\le x\le 2\pi$
In this interval $\left(1\right)$ gives
$x=\pi /8,5\pi /8,9\pi /8,13\pi /8$. $(n=0,1,2,3)$
and $\left(2\right)$ gives $x=2\pi /3,4\pi /3$. (for $n=0,1$)
Thus we get the answer
$x=\pi /8,5\pi /8,2\pi /3,9\pi /8,4\pi /3,13\pi /8$.