Question

sin2x+2√3cos^x-6sinx-6√3cosx=0,на промежутке [pi/2;2*pi].

Answer 1

$sin2x+2sqrt3cos^2x-6sin x-6sqrt3cos x=0\\2sin xcos x-6sin x+2sqrt3cos^2x-6sqrt3cos x=0\2sin x(cos x-3)+2sqrt3cos x(cos x-3)=0\(2sin x+2sqrt3cos x)(cos x-3)=0\begin{cases}2sin x+2sqrt3cos x=0\cos x-3=0;-;pew.;HETend{cases}\2sin x+2sqrt3cos x=0\2sin x=-2sqrt3cos x\sin x=-sqrt3cos x\tgx=sqrt3\x=fracpi3+pi n,;ninmathbb{Z}$
$xinleft[fracpi2;;2piright]Rightarrowfracpi2leqfracpi 3+pi n leq2pi\fracpi6leqpi nleqfrac{5pi}3\frac16leq nleqfrac53\ninmathbb{Z}Rightarrow n=1\x=fracpi3+pi=frac{4pi}3$