Question

cos(x)-sin(x)-2sin(x)*cos(x)=1

Answer 1

Решается введением переменной $t=cosx-sinx$
$cosx-sinx=t \ (cosx-sinx)^2=t^2 \ 1-2sinxcosx=t^2 \ 2sinxcosx=1-t^2$
Подставим в исходное выражение.
$t-(1-t^2)=1 \ t-1+t^2-1=0 \ t^2+t-2=0 \ According.to.Vieta's .formulas .applied. to .quadratic .polynomial: \ t_{1}+ t_{2} =- frac{b}{a} ; t_{1} t_{2} = frac{c}{a} \ t_{1}+ t_{2} =- frac{1}{1} =-1 \ t_{1} t_{2} = frac{-2}{1} =-2 \ Then: t_{1}=-2; t_{2}=1 \ Return:t=cosx-sinx \ cosx-sinx=-2: no.roots:sinx,cosx neq -1 simultaneously \ cosx-sinx=-1 \ auxiliary.angle.method: sqrt{2}/2cosx-sqrt{2}/2sinx=-sqrt{2}/2 \ sin( pi /4)cosx-cos( pi /4)sinx=-sqrt{2}/2$$sin( pi /4)cosx-cos( pi /4)sinx=-sqrt{2}/2 \ sin( pi /4-x)=-sqrt{2}/2 \ x=(-1)^narcsin(-sqrt{2}/2)-( pi /4)+ pi n, n:integer \ x=- pi /4+(-1)^n (5 pi /4)+ pi n$