Question

1+3cosx=sin2x+3sinx
Решите пожалуйста)!

Answer 1

Решение:
1-sin2x=cosx-sinx
(cosx-sinx)²=cosx-sinx
(cosx-sinx)²-(cosx-sinx )=0
(cosx-sinx)(cosx-sinx-1)=0
a) cosx-sinx=0
1-tgx=0
tgx=1
x1=π/4+πn
б) cosx-sinx-1=0
cos²(x/2)-sin²(x/2)-2sin(x/2)cos(x/2)-cos²(x/2)-sin²(x/2)=0
-2sin²(x/2)-2sin(x/2)cos(x/2)=0
2sin(x/2)(sin(x/2)-cos(x/2))=0
sin(x/2)=0
x/2=πn
x2=2πn
sin(x/2)-cos(x/2)=0
tg(x/2)=1
x/2=π/4+πn
x3=π/2+2πn