Question

4sin2x+8(sinx-cosx)=7

Answer 1

$4sin 2x+8(sin x-cos x)=7\ 4sin 2x+8(sin x-cos x)=7(sin^2x+cos^2x)\ 4sin2x+8(sin x-cos x)=7(sin^2x+cos^2x-sin2x+sin2x)=0\ 4sin2x+8(sin x-cos x)=7((sin x - cos x)^2+sin 2x)\ 4sin 2x+8(sin x-cos x)=7(sin x-cos x)^2+7sin 2x\ 7(sin x-cos x)^2-8(sin x-cos x)+3sin 2x=0$
 Пусть $sin x-cos x=t,,(|t| leq sqrt{2} )$, тогда
 $(sin x-cos x)^2=t^2\ 1-sin2x=t^2\ sin2x = 1-t^2$
В результате замены переменных, получаем уравнение
$7t^2-8t+3(1-t^2)=0\ 4t^2-8t+3=0\ D=b^2-4ac= 64-48=46;,, sqrt{D} =4\ t_1= frac{8-4}{8}=0.5$
$t_2= frac{8+4}{8}=1.5notin|t| leq sqrt{2}$

Возвращаемся к замене
$sin x-cos x=0.5\ sqrt{2}sin(x- frac{pi}{4})=0.5|:sqrt{2} \ sin (x- frac{pi}{4})= frac{sqrt{2}}{4} \ x- frac{pi}{4}=(-1)^kcdot arcsin(frac{sqrt{2}}{4})+ pi k,k in Z \ x=(-1)^kcdot arcsin(frac{sqrt{2}}{4})+frac{pi}{4}+ pi k,k in Z$