Sin2x=sin (π/2+x) помогите решить тригонометрическое уравнение

Sin2x = sin (π/2+x)

2sinxcosx = cosx

2sinxcosx - cosx = 0

cosx (2sinx - 1) = 0

1) cosx = 0

x = π/2 + πn, n∈Z

2) 2sinx - 1 = 0

sinx = 1/2

x = (-1) ^n * π/4 + πn, n∈Z

Ответ: x = π/2 + πn, n∈Z и x = (-1) ^n * π/4 + πn, n∈Z