Решите пожалуйста уравнение:sin2x=sin(x+pi/2)

                         sin2x=sin(x+pi/2)
                         sin2x=cosx
                        2sinx*cosx-cosx=0
                        cosx(2sinx-1)=0
           cosx=0                           2sinx-1=0
  x=+-П/2+2Пn, n∈Z                 2sinx=1
                                                  sinx=1/2
                                                 x=(-1)^n *П/6 + Пn, n∈Z