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