(1-tg x )/(1+ tg ^2 x)=2cos2x 

(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

[latex] \frac{1-tgx}{1+tg^2x} = \frac{1- \frac{sinx}{cosx} }{ \frac{1}{cos^2x} } =cos^2x-sinx*cosx[/latex] [latex]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 [/latex] Пусть tg x = t ( t ∈ R) ,  тогда имеем [latex]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} [/latex] Обратная замена [latex]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[/latex]