Question

Помогите решить!!
3sinx-2sinx*cosx-3cosx=2

Answer 1

$3sin x-2sin xcos x-3cos x= 2\ 3sin x - 2sin xcos x - 3cos x = 2sin^2x +2cos^2x\ 3(sin x-cos x)-2sin xcos x-2(sin^2x+cos^2x)=0\ 3(sin x-cos x)-sin2x-2(sin^2x+cos^2x+sin2x-sin2x)=0\ -2(sin x-cos x)^2-3(sin x-cos x)-3sin 2x=0$

Пусть $sin x - cos x = t(|t|<=1)\ 1-sin2x=t^2,,,,= textgreater sin2x=1-t^2$
Получаем
$-2t^2+3t-3(1-t^2)=0\ t^2+3t-3=0\ D=9+12=21\ t_1= frac{-9+ sqrt{21} }{2}\ t_2=frac{-9- sqrt{21} }{2}$

Возвращаемся к замене
$sin x-cos x = frac{-9+ sqrt{21} }{2}\ sqrt{2} sin (x- frac{pi}{4})=frac{-9+ sqrt{21} }{2}\ x=(-1)^{k}cdot arcsin(frac{-9+ sqrt{21} }{2sqrt{2}}) + frac{pi}{4} +pi k,k in Z$