Найдите корни уравнения
sin (3х+4π/3)+cos(7π/6−3х) =−1 на интервале [0;2π]
sin (3х+4π/3)+cos(7π/6−3х) =−1 на интервале [0;2π]
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Answers & Comments
sin(3*x)*cos(4*П/3)+сos(3*x)*sin(4*П/3)+cos(7*П/6)*cos(3*x)+sin(7*П/6)*sin(3*x)=-1
sin(3*x)*(-1/2)+cos(3*x)*(-√3/2)+(-√3/2)*cos(3*x)+(-1/2)*sin(3*x)=-1
-sin(3*x)-√3*cos(3*x)=-1
sin(3*x)+√3*cos(3*x)=1 возведём в степень обе части уравнения
sin^2(3*x)+2*√3*sin(3*x)*cos(3*x)+3*cos^2(3*x)=1
sin^2(3*x)+cos^2(3*x)+2*√3*sin(3*x)*cos(3*x)+2*cos^2(3*x)=1
1+2*√3*sin(3*x)*cos(3*x)+2*cos^2(3*x)=1
2*√3*sin(3*x)*cos(3*x)+2*cos(3*x)=0
2*cos(3*x)*(√3*sin(3*x)+cos(3*x))=0
2*cos(3*x)=0
cos(3*x)=0 3*x=П/2+П*n x=П/6+П*n/3, nЄZ
x=П/6; П/2; 5*П/6; 7*П/6; 3*П/2; 11*П/6 на промежутке [0; 2*П]
√3*sin(3*x)+cos(3*x)=0
сos(3x)=-√3*sin(3*x) разделим уравнение на sin(3*x)
ctq(3*x)=-√3 3*x=5*П/6+П*k x=5*П/18+П*k/3, kЄZ
x=5*П/18; 11*П/18; 17*П/18; 23*П/18; 29*П/18; 35*П/18 на интервале
[0; 2*П]