Решите : 1) 2cos(x/3-π/3)=√3 2) cos(π/2+2x)=1 3) sin(2x-π/4)=0

1) 2cos(x/3-π/3)=√3
cos(x/3-π/3)=√3/2
x/3-π/3=-π/6+2πn,n∈Z U x/3-π/3=π/6+2πk,k∈Z
x/3=π/6+2πn,n∈Z U x/3=π/2+2πk,k∈Z
x=π/2+6πn,n∈Z U x=3π/2+6πk,k∈Z
2) cos(π/2+2x)=1
-sin2x=1
sin2x=-1
2x=-π/2+2πn,n∈Z
x=-π/4+πn,n∈Z
3) sin(2x-π/4)=0
2x-π/4=πn,n∈Z
2x=π/4+πn,n∈Z
x=π/8+πn/2,n∈Z