4sin(2x-п/5)*cos(2x-п/5)=1

Формула 2sinα·cosα=sin2α 2sin(4x-(2π/5))=1 sin(4x-(2π/5))=1/2 4x+(2π/5)=(π/6)+2πk, k∈ Z    или    4x=(2π/5)=(5π/6)+2πn, n∈ Z  4x=(2π/5)+(π/6)+2πk, k∈ Z    или    4x=(2π/5)+(5π/6)+2πn, n∈ Z  x=(17π/120)+(π/2)·k, k∈ Z    или    x=(37π/120)+(π/2)·n, n∈ Z  О т в е т.(17π/120)+(π/2)·k,  (37π/120)+(π/2)·n,    k, n∈ Z