Question

Найти корни уравнения sin (2x - П/2) = -1/2, принадлежащие полуинтервалу (0; 3П/2].

Answer 1

sin (2x - П/2) = -1/2
2x-π/2=-(-1)^n*π/6+πk
2x=π/2-π/6=π/3  x=π/6
2x=π/2+7π/6=5π/3  x=5π/6