cos2x=(sqrt2)(cosx-sinx)
(cosx-sinx)*(cosx+sinx)=sqrt(2)*(cosx-sinx)
(cosx-sinx)*(cosx+sinx)-sqrt(2)*(cosx-sinx)=0
(cosx-sinx)*(cosx+sinx-sqrt(2))=0
cosx-sinx=0 или cosx+sinx-sqrt(2)=0
cosx-sinx=0
tgx=1
[b]x=(π/4)+πk, k ∈ Z[/b]
или
cosx+sinx-sqrt(2)=0
cosx+sinx=sqrt(2)
по формулам приведения: [blue]cosx=sin((π/2)-x)[/blue]
sin((π/2)-x)+sinx=sqrt(2)
2sin(π/4)* cos((π/4)-x)=sqrt(2), так как [blue]sin(π/4)=sqrt(2)/2[/blue]
cos((π/4)-x)=1
cos((π/4)-x)=cos(x-(π/4))
cos(x-(π/4)=1
x-(π/4)=2πn, n ∈ Z
[b]x=(π/4) + 2πn, n ∈ Z[/b]
О т в е т.[b](π/4)+πk, k ∈ Z;(π/4) + 2πn, n ∈ Z[/b]
наибольший отрицательный
x=(π/4)-π=-3π/4