Question

Найдите корни уравнения sin(3x-Π/6)=1/2, принадлежащие промежутку [-Π;Π)

Answer 1

..корни уравнения sin(3x-Π/6)=1/2,
3x=π/6+(-1)^n*π/6+πn
x=π/18+(-1)^n*π/18+πn/3
n=0  x=π/18
n=1  x=π/18+π/3-π/18=π/3
n=-1  x=π/18+π/3+π/3=2π/3+π/18=13π/18
n=2   x=π/18+2π/3+2π/18=π/6+2π/3=5π/6
n=-2  x=π/18-2π/3-π/9=-π/18-2π/3=-11π/3<-π
n=3   x=π/18+π-π/3=19π/18-6π/18=13π/18
n=4  x=π/18+4π/3+4π/18=29π/18 бл. π

π/3  13π/18  5π/6