Решите пожалуйста: 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)=√3cos(x/3-π/3)=√3/2x/3-π/3=-π/6+2πn,n∈Z U x/3-π/3=π/6+2πk,k∈Zx/3=π/6+2πn,n∈Z U x/3=π/2+2πk,k∈Zx=π/2+6πn,n∈Z U x=3π/2+6πk,k∈Z2) cos(π/2+2x)=1-sin2x=1sin2x=-12x=-π/2+2πn,n∈Zx=-π/4+πn,n∈Z 3) sin(2x-π/4)=0 2x-π/4=πn,n∈Z2x=π/4+πn,n∈Zx=π/8+πn/2,n∈Z