Question

Найдите решения уравнения sin(3x-pi/4)=0 на промежутке [0;pi]

Answer 1

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Answer 2

sin(3x-π/4)=0
3x-π/4=πk
3x=π/4+πk
x=π/12+πk/3
0≤x≤π
0≤π/12+πk/3≤π
-π/12≤πk/3≤π-π/12
-π/12:π/3≤k≤11π/12:π/3
-1/4≤k≤11/4
k={0;1;2}
x1=π/12
x2=π/12+π/3=(π+4π)/12=5π/12

x3=π/12+2π/3=(π+8π)/12=9π/12=3π/4