Odpowiedź: (x = -π/12 + 2kπ/3 v x = -π/4 + 2kπ/3), k ∈ C
Rozwiązanie:
sin (-3x)=√2/2
sin a = √2/2 <=> (a = π/4 + 2kπ v a = 3π/4 + 2kπ), k ∈ C
Zatem: (-3x = π/4 + 2kπ v -3x = 3π/4 + 2kπ), k ∈ C
(x = -π/12 + 2kπ/3 v x = -π/4 + 2kπ/3), k ∈ C
