2 корень из 2 sin(х+П/6)-корень 3 cos 2х= корень из 2 cosх-корень из 3

Question

eugene_greek · Accepted Answer

2*sqrt(2)*sin(x+pi/6)-sqrt(3)*cos 2x=sqrt(2)*cos x-sqrt(3)
2*sqrt(2)*(sin pi/6*cos x+cos pi/6*sin x)-sqrt(3)*(cos^2 x-sin^2 x)=sqrt(2)*cos x-sqrt(3)
2*sqrt(2)*(1/2*cos x+sqrt(3)/2*sin x)-sqrt(3)*(cos^2 x-sin^2 x)=sqrt(2)*cos x-sqrt(3)
sqrt(2)*(cos x+sqrt(3)*sin x)-sqrt(3)*(cos^2 x-sin^2 x)=sqrt(2)*cos x-sqrt(3)
sqrt(2)*cos x+sqrt(6)*sin x-sqrt(3)*cos^2 x-sqrt(2)*sin^2 x=sqrt(2)*cos x-sqrt(3)
sqrt(6)*sin x-sqrt(3)*cos^2 x-sqrt(2)*sin^2 x=-sqrt(3)
sqrt(6)*sin x-sqrt(3)*(1-sin^2 x)-sqrt(2)*sin^2 x=-sqrt(3)
sqrt(6)*sin x-sqrt(3)-sqrt(3)*sin^2 x-sqrt(2)*sin^2 x=-sqrt(3)
sqrt(6)*sin x-sqrt(3)*sin^2 x-sqrt(2)*sin^2 x=0|*(-1)
sin^2 x(sqrt(3)+sqrt(2))-sqrt(6)*sin x=0
sin x(sin x*(sqrt(3)+sqrt(2))-sqrt(6))=0
sin x=0
x=pi*n
sin x*(sqrt(3)+sqrt(2))-sqrt(6)=0
sin x*(sqrt(3)+sqrt(2))=sqrt(6)
sin x=sqrt(6)/(sqrt(3)+sqrt(2))
x=arcsin (sqrt(6)/(sqrt(3)+sqrt(2)))+2*pi*n