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Verified Answer :

Julik

\(sin^2 \alpha +cos^2 \alpha =1 \\ sin^2x=1-cos^2x \\ \\ cos(2 \alpha )=cos^2 \alpha -sin^2 \alpha \\cos(4x)=cos^2(2x)-sin^2(2x) \\ \\ \\ \\ cos (4x)-cos(2x)=0 \\cos^2(2x)-sin^2(2x)-cos(2x)=0 \\ \cos^2(2x)-(1-cos^2(2x))-cos(2x)=0 \\ cos^(2x)+cos^2(2x)-cos(2x)-1=0 \\ 2cos^2(2x)-cos2x-1=0 \\ t=cos(2x) \\\)
\(2t^2-t-1=0 \\ t_1= \frac{-(-1)- \sqrt{(-1)^2-4\cdot2\cdot(-1)} }{2 \cdot2} = \frac{1- \sqrt{1+8} }{4} = \frac{1- \sqrt{9} }{4} = \frac{1-3}{4} = \frac{-2}{4} =- \frac{1}{2} \\ t_1= \frac{-(-1)+ \sqrt{(-1)^2+4\cdot2\cdot(-1)} }{2 \cdot2} = \frac{1+\sqrt{1+8} }{4} = \frac{1+ \sqrt{9} }{4} = \frac{1+3}{4} = \frac{4}{4} =1 \\\)

\(cos(2x)=- \frac{1}{2 } \\ 2x= \pm \frac{ 2\pi }{3}+ 2 \pi k,~~k \in Z \\ x= \pm \frac{ \pi }{3}+ \pi k,~~k \in Z \\ \\ cos(2x)=1 \\ 2x=2 \pi k,~~k \in Z \\ x=\pi k,~~k \in Z \\ \\ x\in[0, 2 \pi ) \\ x= \pm \frac{ \pi }{3}+ \pi k,~~k \in Z \\ x_1= \frac{ \pi }{3} ,~~~~~x_2=\frac{ \pi }{3}+ \pi = \frac{4 \pi }{3} ,~~~~~ \\ x_3=-\frac{ \pi }{3}+ \pi = \frac{2 \pi }{3} ~~~~~~x_4=-\frac{ \pi }{3}+ 2\pi = \frac{5 \pi }{3} \\ x=\pi k,~~k \in Z \\ x_4=0 ~~~~ x_5= \pi\)

cos 4 x - cos 2 x = 0
2cos^2 2x-1 - cos 2x = 0

2u^2 -u -1 = 0

(u-1)(2u+1) = 0
u = 1; u = -1/2

cos 2x = 1 ;
cos 2x = -1/2
2x = 0 ;
 2x = pi-pi/3,
2x = pi + pi/3
x = 0 + npi; x = pi/3 + npi, x = 2pi/3 + npi
x = 0, pi; x = pi/3, 4pi/3,
x = 2pi/3, 5pi/3 \
so solutions will be
 x = 0, pi/3, 2pi/3, 4pi/3, 5pi/3, pi
hope it helps

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