2sin²2x-1=cos2x
2(1-cos²2x)-1=cos2x
2-2cos²2x-1-cos2x=0
2cos²2x+cos2x-1=0
Пусть cos2x=t (|t|≤1) имеем
2t²+t-1=0
D=b²-4ac=1+8=9
t1=(-1+3)/4=1/2
t2=(-1-3)/4=-1
Возвращаемся к замене
cos2x=1/2
2x=±π/3+πn,n ∈ Z
x=±π/6 + πn/2, n∈ Z
cos2x=-1
2x=π+2πn,n ∈ Z
x=π/2+πn,n ∈ Z