Question

Система
{
x-y=13п/2 ;
3cos^x-12cosy+4=0

Answer 1

x=13π/2+y
3cos²(13π/2+y)-12cosy+=0
3sin²y-12cosy+4=0
3-3cos²y-12cosy+4=0
cosy=a
3a²+12a-7=0
D=144+84=228
a1=(-12-2√57)/6⇒cosy=(-12-2√57)/6<-1 нет решения
a2=(-12+2√57)/6⇒cosy=(-12+2√57)/6
y1=-arccos(2√57-12)/6⇒x1=13π/2-arccos(2√57-12)/6
y2=+arccos(2√57-12)/6⇒x2=13π/2+arccos(2√57-12)/6