2(2cos²2x-1)-1-4cos³2x+3cos2x=0
4cos²2x-2-1-4cos³2x+3cos2x=0
4cos³2x-4cos²2x-3cos2x+3=0
t=cos2x, 4t³-4t²-3t+3=0, 4t²(t-1)-3(t-1)=0, (t-1)(4t²-3)=0
t-1=0,t=1, cos2x=1, 2x=2πn, x=πn, n∈Z
4t²-3=0, t²=cos²2x=(1+cos4x)/2, ⇒ 2(1+cos4x)-3=0, 2cos4x=1, cos4x=1/2, 4x=±π/3+2πk,
x=±π/12+πk/2, k∈Z
Ответ: x=πn, n∈Z, x=±π/12+πk/2,k∈Z