Question

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

• A. $0$
• B. $4$
• C. $6$
• D. $10$

Answer 1

The correct option is C $6$

${\cos }^{5}x+{\cos }^5(x+\frac{2\pi }{3})+{\cos }^5(x+\frac{4\pi }{3})=0$

$\Rightarrow {(e^{ix}+e^{-ix})}^5+{(\omega e^{ix}+{\omega }^2{e}^{-ix})}^5+{({\omega }^2{e}^{ix}+\omega e^{-ix})}^5=0$

$\Rightarrow {}^5{C}_0{e}^{5ix}+{}^5{C}_1{e}^{3ix}+{}^5{C}_2{e}^{ix}+\cdots +{}^5{C}_5{e}^{-5ix}+{}^5{C}_0{\omega }^2{e}^{5ix}+{}^5{C}_1{e}^{3ix}+{}^5{C}_2\omega e^{ix}+\cdots +{}^5{C}_5\omega e^{-5ix}+{}^5{C}_0\omega e^{5ix}+{}^5{C}_1{e}^{3ix}+{}^5{C}_2{\omega }^2{e}^{ix}+\cdots +{}^5{C}_5{\omega }^2{e}^{-5ix}=0$

$\Rightarrow 3\cdot {}^5{C}_1{e}^{3ix}+3\cdot {}^5{C}_4{e}^{-3ix}=0$
$\Rightarrow \cos 3x=0$
$\Rightarrow x=(2n+1)\frac{\pi }{6},$ $n\in Z$
$x\in \{\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\frac{7\pi }{6},\frac{3\pi }{2},\frac{11\pi }{6}\}$