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 equal to.

$$\begin{gathered} {(e^{ix}+e^{-ix})}^5+{(\omega e^{ix}+{\omega }^2{e}^{-ix})}^5+{({\omega }^2{e}^{ix}+\omega e^{-ix})}^5=0 \\ {\Rightarrow }^5{C}_0{e}^{5ix}{+}^5{C}_1{e}^{3ix}{+}^5{C}_2{e}^{-ix}+\mathrm{.......}{+}^5{C}_5{e}^{-5ix} \\ {+}^5{C}_0{\omega }^2{e}^{5ix}{+}^5{C}_1{e}^{3ix}{+}^5{C}_2\omega e^{ix}+\mathrm{.......}{+}^5{C}_5\omega e^{-5ix} \\ {+}^5{C}_0\omega e^{5ix}{+}^5{C}_1{e}^{3ix}{+}^5{C}_2{\omega }^2{e}^{ix}+\mathrm{.......}{+}^5{C}_5{\omega }^2{e}^{-5ix} \end{gathered}$$

$$\begin{gathered} =0 \\ {\Rightarrow }^5{C}_1{e}^{3ix}{+}^5{C}_4{e}^{-3ix}=0 \Rightarrow 2\cos 3x=0 \Rightarrow \cos 3x=0 \\ \Rightarrow 3x=(2n+1)\frac{\pi }{2},n\in I \end{gathered}$$

$x=\{\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\frac{7\pi }{6},\frac{3\pi }{2},\frac{11\pi }{6}\}$

$\therefore$ Total number of solution=6.