2 sin(3x-Π/4)+1=0 хелп плз!!!

Решение 2 sin(3x-Π/4)+1=0   sin(3x-Π/4) = - 1/2 3x - π/4 = (-1)^n arcsin(- 1/2) + πk, k ∈ Z 3x - π/4 = (-1)^(n+1) arcsin(1/2) + πk, k ∈ Z 3x - π/4 = (-1)^(n+1) *(π/6) + πk, k ∈ Z 3x  = (-1)^(n+1) *(π/6) + π/4 + πk, k ∈ Z x  = (-1)^(n+1) *(π/18) + π/12 + πk/3, k ∈ Z