Cos4x=2cos²2x-1
2cos²x=1+cos2x
2cos²2x-1+1+cos2x=1
2cos²2x+cos2x-1=0
Пусть cos2x=t (|t|≤1)
2t²+t-1=0
a=2;b=1;c=-1
D=b²-4ac=1-4*2*(-1)=9
√D=3
t₁=(-b+√D)/2a=(-1+3)/4=1/2
t₂=(-b-√D)/2a=(-1-3)/4=-1
замена
cos2x=1/2
2x=±π/3+2πn, n € Z
x₁=±π/6+πn, n € Z
cos2x=-1
2x=π+2πn, n € Z
x₂=π/2+πn, n € Z