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

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+4 П/12=5 П/12∈[0; 4]

x3=П/12+П*2/3=П/12+8 П/12=9 П/12∈[0; 4]

x3=П/12+П*3/3=П/12+12 П/12=13 П/12∈[0; 4]

x4=П/12+П*4/3=П/12+16 П/12=17 П/12∉[0; 4]

Ответ: 5 П/12; 9 П/12; 13 П/12

Sin3x=cos3x|:cos 3x

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