Question

решите уравнение √2sin2x+1=0

Answer 1

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

Answer 2

спасибки