2*cos^2(4x)-6*cos^2(2x)+1=0

решение:
2*cos(4x)-6*cos(2x)+1=0
2(2cos2x-1)-6*cos(2x)+1=0
cos2x = y
2(2y-1)-6y+1=0
2(4y-4y+1)-6y+1=0
8y-8y+2-6y+1=0
8y-14y+3=0
D=14-4*8*3 = 196-96 = 100
y1=(14+√100)/(2*8) = 24/16 = 3/2
y2=(14-√100)/(2*8) = 4/16 = 1/4
y = cos2x =>
0≤y≤1 =>
y = 1/4
cos2x = 1/4
cos2x1 = 1/2; cos2x2 = -1/2
2x1 = π/3
2x2 = 2π/3
x1 = π/6
x2 = π/3