Cos8x=cos4x
cos(2*4x)=cos4x
cos² 4x - sin² 4x - cos4x=0
cos² 4x - (1-cos² 4x) - cos4x=0
cos² 4x - 1 + cos² 4x - cos4x=0
2cos² 4x - cos4x - 1=0
y=cos4x
2y² - y - 1=0
D=1 +8=9
y₁=(1-3)/4= -2/4= -1/2
y₂=(1+3)/4=1
При у= -1/2
cos4x= -1/2
4x=(+/-) (2π/3)+2πk, k∈Z
x=(+/-) (π/6) + (π/2)k, k∈Z
При у=1
cos4x=1
4x=2πn, n∈Z
x=(π/2)n, n∈Z
Ответ: (+/-) (π/6)+(π/2)k, k∈Z;
(π/2)n, n∈Z.