Помогите решить sin (8*x-pi/6)=sinx
2sin(7x/2-π/12)cos(6x/2-π/12)=0
sin(7x/2-π/12)=0
7x/2-π/12=πn,n∈z
7x/2=π/12+πn,n∈z
x=π/42+2πn/7,n∈z
sin(9x/2-π/12)=0
9x/2-π/12=πk,k∈z
9x/2=π/12+πk,k∈z
x=π/54+2πk/9,k∈z
Оцени ответ
Вход
Регистрация
Задать вопрос
2sin(7x/2-π/12)cos(6x/2-π/12)=0
sin(7x/2-π/12)=0
7x/2-π/12=πn,n∈z
7x/2=π/12+πn,n∈z
x=π/42+2πn/7,n∈z
sin(9x/2-π/12)=0
9x/2-π/12=πk,k∈z
9x/2=π/12+πk,k∈z
x=π/54+2πk/9,k∈z