Question

(1-tg x )/(1+ tg ^2 x)=2cos2x 

Answer 1

(1-tg x)/(1+ tg² x)=2cos2x 
(cosx-sinx)/cosx*cos²x=2(cosx-sinx)(cosx+six)
(cosx-sinx)cosx=2(cosx-sinx)(cosx+six)
1) cosx-sinx=0
sinx=cosx
tgx=1
x=π/4+πn
2) cosx=2(cosx+six)
2sinx=-cosx
2tgx=-1
tgx=-1/2
x=arctg(-1/2)+πn
x=-arctg1/2+πn

Answer 2

$frac{1-tgx}{1+tg^2x} = frac{1- frac{sinx}{cosx} }{ frac{1}{cos^2x} } =cos^2x-sinx*cosx$

$cos^2x-sinx*cosx=2cos2x \ cos^2x-sin*cosx=2cos^2x-2sin^2x \ 2sin^2x-sinx*cosx-cos^2x=0|:cos^2x \ 2tg^2x-tgx-1=0$

Пусть tg x = t ( t ∈ R) ,  тогда имеем

$2t^2-t-1=0 \ D=b^2-4ac=(-1)^2-4*2*(-1)=9 \ sqrt{D} =3 \ t_1= frac{-b+ sqrt{D} }{2a} = frac{1+3}{4} =1 \ t_2= frac{-b- sqrt{D} }{2a} = frac{1-3}{4} =- frac{1}{2}$

Обратная замена

$tgx=1 \ x_1=arctg1+ pi n \ x_1= frac{ pi }{4} + pi n \ tgx=- frac{1}{2} \ x_2=-arctg frac{1}{2} + pi n$