Question

$\text{If}{\cos }^{5}x+{\cos }^5(x+\frac{2\pi }{3})+{\cos }^5(x+\frac{4\pi }{3})=0$,
then the number of solution(s) in $[0,2\pi ]$ is

Answer 1

${\cos }^{5}x+{\cos }^5(x+\frac{2\pi }{3})+{\cos }^5(x+\frac{4\pi }{3})=0\Rightarrow (e^{ix}+e^{-ix}{)}^5+(\omega e^{ix}+{\omega }^2{e}^{-ix}{)}^5+({\omega }^2{e}^{ix}+\omega e^{-ix}{)}^5=0\Rightarrow {}^5{C}_0{e}^{i5x}+{}^5{C}_1{e}^{i3x}+{}^5{C}_2{e}^{ix}+\dots {}^5{C}_5{e}^{-i5x}+{}^5{C}_0{\omega }^2{e}^{i5x}+{}^5{C}_1{e}^{i3x}+{}^5{C}_2\omega e^{ix}+\dots {}^5{C}_5\omega e^{-i5x}+{}^5{C}_0\omega e^{i5x}+{}^5{C}_1{e}^{i3x}+{}^5{C}_2{\omega }^2{e}^{ix}+\dots {}^5{C}_5{\omega }^2{e}^{-i5x}=0\Rightarrow {}^5{C}_1{e}^{i3x}+{}^5{C}_4{e}^{-i3x}=0\Rightarrow 2\cos 3x=0\Rightarrow \cos 3x=0\Rightarrow 3x=(2n+1)\frac{\pi }{2},n\in Ix=\{\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6}\frac{7\pi }{6},\frac{3\pi }{2},\frac{11\pi }{6}\}$
Total number of solutions $6$.