Find the number of solutions of sin x+sin 2x+sin 3x=cos x+cos 2x+cos 3x in the interval [0,2π].

Group the terms: $\sin x+\sin3x=2\sin2x\cos x$ and $\cos x+\cos3x=2\cos2x\cos x$. The equation becomes $\sin2x(1+2\cos x)=\cos2x(1+2\cos x)$, that is $(1+2\cos x)(\sin2x-\cos2x)=0$. Case 1: $\cos x=-\tfrac12$ gives $x=\tfrac{2\pi}{3},\tfrac{4\pi}{3}$, two solutions. Case 2: $\tan2x=1$ gives $2x=\tfrac\pi4+k\pi$, so $x=\tfrac\pi8+\tfrac{k\pi}{2}$ for $k=0,1,2,3$ in $[0,2\pi]$, four solutions: $\tfrac\pi8,\tfrac{5\pi}8,\tfrac{9\pi}8,\tfrac{13\pi}8$. None coincides with the first set, since $\cos$ of these angles is not $-\tfrac12$. A numerical sign-change count over a fine grid also gave $6$ roots. The total is $2+4=6$. Therefore the required count of solutions in $[0,2\pi]$ is $6$, as obtained both by factorisation and by the numerical check, with no root double counted or lying on the boundary of the interval.