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