Question

The number of values of $x$ between $0$ and $2\pi$ that satisfies the equation $\sin x+\sin 2x+\sin 3x=\cos x+\cos 2x+\cos 3x,$, are

Answer 1

We have,

$\begin{array}{c}\sin x+\sin 2x+\sin 3x=\cos x+\cos 2x+\cos 3x\\ 2\sin 2x\cos x+\sin 2x=2\cos 2x\cos x+\cos 2x\\ \sin 2x\left(2\cos x+1\right)=\cos 2x\left(2\cos x+1\right)\\ \left(\sin 2x-\cos 2x\right)\left(2\cos x+1\right)=0\\ \sin 2x=\cos 2xOR\cos x=-\frac{1}{2}\\ \tan 2x=1x=\frac{2\pi }{3},\frac{4\pi }{3}\\ OR\\ 2x=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4},\frac{13\pi }{4}\\ x=\frac{\pi }{8},\frac{5\pi }{4},\frac{9\pi }{4},\frac{13\pi }{4}\\ =\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},\frac{13\pi }{8},\frac{2\pi }{3},\frac{4\pi }{3}\end{array}$

Total $6$ values given this equation.