4cos²2x +2sin2x = 1;
4(1- sin²2x) +2sin2x =1; * * * cos²α +sin²α =1 ; α=2x * * *
4sin²2x -2sin2x -3 = 0 ;
замена : t =sin2x , |t| ≤ 0;
4t² -2t -3 =0;
D/4 =1+4*3 =13;
t₁ =(1+√13)/4 >1 не решения ;
t₂ =(1- √13)/4 .
sin2x = (1-√13)/4;
2x = (-1)^(k+1)arcsin(√13 -1)/4 +πk ,k∈Z .
x = (-1)^(k+1)*1/2*(arcsin(√13 -1)/4) +π/2*k , k∈Z .