Question

2cosx×cos4x+cosx-2cos4x -1=0​

Answer 1

$2Cosx\cdot Cos4x+Cosx-2Cos4x-1=0\\\\(2Cosx\cdot Cos4x-2Cos4x)+(Cosx-1)=0\\\\2Cos4x(Cosx-1)+(Cosx-1)=0\\\\(Cosx-1)(2Cos4x+1)=0\\\\\left\{\begin{array}{ccc}Cosx-1=0\\2Cos4x+1=0\end{array}\right \\\\\\\left\{\begin{array}{ccc}Cosx=1\\2Cos4x=-1\end{array}\right\\\\\\\left\{\begin{array}{ccc}Cosx=1\\Cos4x=-\dfrac{1}{2} \end{array}\right \\\\\\1) \ Cosx=1\\\\\boxed{x=2\pi n, \ n\in Z}$

$2) \ Cos4x=-\dfrac{1}{2}\\\\4x=\pm arcCos\Big(-\dfrac{1}{2}\Big)+2\pi n,n\in Z\\\\4x=\pm \dfrac{2\pi }{3}+2\pi n,n\in Z\\\\\boxed{x=\pm \dfrac{\pi }{6}+\dfrac{\pi n }{2} ,n\in Z}$