Question

Решить уравнение: 2sin(3x-п/4)=-√2

Answer 1

sin(3x - π/4) = -√2/2
3x - π/4 = (-1)^n*arcsin(-√2/2) + πn, n∈Z
3x - π/4 = (-1)^(n+1)*arcsin(√2/2) + πn, n∈Z
3x - π/4 = (-1)^(n+1)*π/4) + πn, n∈Z
3x = (-1)^(n+1)*(π/4) + π/4+ πn, n∈Z
x = (-1)^(n+1)*(π/12) + π/12+(πn)/3, n∈Z