Question

4 - sinxcosx + 4cosx = 4sinx

Answer 1

$4-sin xcos x+4cos x=4sin x \ \ 4(cos^2x+sin^2x)-sin xcos x+4cos x-4sin x=0 \ 4(cos^2+sin^2-2sin xcos x+2sin xcos x)-sin xcos x+4(cos x-sin x)=0$
$4((cos x-sin x)^2+2sin xcos x)-sin xcos x+4(cos x-sin x)=0 \ 4(cos x-sin x)^2+7sin xcos x+4(cos x-sin x)=0$
 Пусть $(sin x-cos x)=t(|t| leq sqrt{2} )$, тогда $1-2sin xcos x=t^2 \ sin xcos x= frac{1-t^2}{2}$
Получаем
$4t^2+7 frac{1-t^2}{2} +4t=0 \ 8t^2+7-7t^2+8t=0 \ t^2+8t+7=0 \ D=64-28=36 \ t_1=-1 \ t_2=-7$
Корень t=-7 - не удовлетворяет условие при |t|≤√2

Возвращаемся к замене
$cos x-sin x=-1 \ sqrt{2} sin (x- frac{ pi }{4} )=-1 \ x-frac{ pi }{4}=(-1)^k^+^1cdot frac{ pi }{4}+ pi k,k in Z \ x=(-1)^k^+^1cdot frac{ pi }{4}+frac{ pi }{4}+ pi k,k in Z$