ПОМОГИТЕ ПОЖАЛУЙСТА СРОЧНО 1+sin2x-sinx=cosx нужно решить ...

Vasy

1+sin2x-sinx=cosx

cos^2(x)+sin^2(x)+2sinxcosx-sinx-cosx=0

(sinx+cosx)^2-(sinx+cosx)=0

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

 

sinx+cosx=0|:cosx не равный 0    или    sinx+cosx-1=0

tgx+1=0                                                sinx+cosx=1

tgx=-1                                                  sqrt{2}cos(П/4-x)=1

x=-П/4+Пn, n принадлежит N               cos(П/4-x)=1/sqrt{2}

                                                           cos(П/4-x)=sqrt[2}/2

                                                           П/4-x=(+-)П/4+2Пn, n N

                                                           -x=(+-)П/4-П/4+2Пn, n N

                                                            x=(-+)П/4+П/4-2Пn, n N