Решите уравнение.
sin(3x+3)=cos(x-1)

sin(3x+3)-sin(π/2-(x-1))=0
Применим формулу sina-sinb=2sin(a-b)/2*cos(a+b)2
[3x+3-(п/2-(х-1)]/2=(3х+3-п/2+х-1)/2=(4x+2-п/2)/2=2х+1-п/4
[3x+3+(п/2-(х-1)]/2=(3х+3+п/2-х+1)/2=(2x+4+п/2)/2=x+2+п/4
2sin(2x+1-π/4)cos(x+2+π/4)=0
sin(2x+1-π/4)=0
2x+1-π/4=πn
2x=π/4-1+πn
x=π/8-1/2+πn/2
cosx(x+2+π/4)=0
x+2+π/4=π/2+πn
x=π/4-2+πn