Question

ПОМОГИТЕ, СРОЧНО!!!!!
cos2x=sinx(sinx-cosx)

Answer 1

cos2x=sinx-cosx

cos²x-sin²x+(cosx-sinx)=0
(cosx-sinx)(cosx+sinx)+(cosx-sinx)=0
(cosx-sinx)(cosx+sinx+1)=0

cosx=sinx => x=π/4+πn

cosx+sinx=-1
sin(x+π/4)=-1/√2
x+π/4=(-1)^(n+1) π/4+πk
x=-π/4+(-1)^(n+1) π/4+πk

Answer 2

Cos^2x-sin^2x-sin^2x+sin x cos x=0;
cos^2 x-2sin^2 x+sin x cos x=0; разделим все на cos^2x не =0;
1-2tg^2х+tg x=0;
2tg^2x-tg x-1=0;
D=9; tgx=-1/2;tgx=1;x=-arctg 1/2+Пn;x=П/4+ пn