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