Question

Найти корни уравнения sin3x=cos3x, принадлежащие отрезку [0;4]

Answer 1

sin3x=cos3x|:cos3x≠0
tg3x=1
3x=П/4+Пn, n∈Z
x=П/12+Пn/3, n∈Z
[0;4]
x1=П/12+П*0/3=П/12∉[0;4]
x2=П/12+П*1/3=П/12+4П/12=5П/12∈[0;4]
x3=П/12+П*2/3=П/12+8П/12=9П/12∈[0;4]
x3=П/12+П*3/3=П/12+12П/12=13П/12∈[0;4]
x4=П/12+П*4/3=П/12+16П/12=17П/12∉[0;4]
Ответ: 5П/12; 9П/12; 13П/12

Answer 2

sin3x=cos3x|:cos3x
tg3x=1
3x=π/4+πk,k∈z
x=π/12+πk/3,k∈z
принадлежат [0;4]
x1=π/12
x2=π/12+π/3=5π/12
x3=π/12+2π/3=9π/12
x4=π/12+π=13π/12
Ответ π/12;5π/12;9π/12;13π/12