Решите уравнение sin(3x+п/4)=1. укажите количество корней уравнения,принадлежащих отрезку [0,2п]

Sin(3x+π/4)=1
(sinπ/2=1)
3x+π/4=π/2+2k.π
3x=π/4+2k.π
x=π/12+2k.π/3 , k∈Z
k=0: x=π/12
k=1:x=π/12+2π/3=9π/12=3π/4
k=2: x=π/12+4π/3=17π/12
k=3: x=π/12+6π/3=π/12+2π - yže net v intervale /0,2π
Otvet: x=π/12, x=3π/4, x=17π/12