Ответ: x∈(-π/3 +2πk; π +2πk) k∈Z

Объяснение:

sin(x+π/6) =-1/2  => x1+π/6 =-π/6 +2πk   или x2+π/6 =-5π/6 +2πk

=> x1=-π/3+2πk         x2=-π+2πk    k∈Z

=> x∈(-π/3 +2πk; π +2πk) k∈Z