6. sin3x=cos3x
tg3x=1
3x=π/4+πk, k∈Z
x=π/12+πk/3, k∈Z
[0;4]
x₁=π/12+π*0/3=π/12∈[0;4]
x₂=π/12+π*1/3=π/12+4π/12=5π/12∈[0;4]
x₃=π/12+π*2/3=π/12+8π/12=9π/12∈[0;4]
x₄=π/12+π*3/3=π12+12π/12=13π/12∈[0;4]
x₅=π/12+π*4/3=π/12+16π/12=17π/12∉[0;4]
Ответ. 5π/12; 9π/12; 13π/12.
