решите уравнение √2sin2x+1=0

спасибки

√2sin2x + 1 = 0sin2x = -1/√22x = (-1)^(n)arcsin(- 1/√2) + πn, n∈Z2x = (-1)^(n+1)arcsin(1/√2) + πn, n∈Z2x = (-1)^(n+1)(π/4) + πn, n∈Zx = (-1)^(n+1)(π/8) + πn/2, n∈Z