sin(2x+п/2) = cos (x+п/2) + sin (x + п/2)

sin(π/4 - x)*sin(π/4 - x/2)*sin(x/2) = 0Sin(2x+п/2) = cos (x+п/2) + sin (x + п/2)cos2x = -sinx + cosx4sin(π/4 - x)*sin(π/4 - x/2)*sin(x/2) = 01)  sin(π/4 - x) = 0π/4 - x = πn, n∈Z- x = πn - π/4, n∈Z  x1 = π/4 - πn , n∈Z2)  sin(π/4 - x/2) = 0π/4 - x/2 = πk, k∈Z - x/2 = πk - π/4 , k∈Z x2  = π/2 - 2πk, k∈Z3)  sin(x/2) = 0(x / 2) = πm, m∈Zx3 =  2πm, m∈ZОтвет:  x1 = π/4 - πn , n∈Z;    x2  = π/2 - 2πk, k∈Z;  x3 =  2πm, m∈Z