2а)
nєZ.
xє[2π/3 + 2πn; π+2πn], nєZ.
2b)
nєZ.
xє[π/4 +πn; π/2+πn], nєZ.
3.
найдем нули функции
соs(x)=0 при х1=π/2 +2πn, nєZ,
x2=3π/2 +2πn, nєZ.
2cos(x)+1=0
cos(x)=-1/2
x1=2π/3 +2πn, nєZ,
x2=4π/3+2πn, nєZ.
___o_____o_____o_____o____
..+..π/2...-...2π/3..+..4π/3...-...3π/2..+.
xє(π/2+2πn;2π/3+2πn)U(4π/3+2πn;3π/2+2πn), nєZ.
-64;32;-16;
b1=-64;b2=32
q=b2/b1=32/(-64)=-1/2
|q|<1
S=b1/(1-q)=(-64)/(1+1/2)=-64*2/3=
-128/3
6х + y = 23
y = 23 - 6x
y = 23
23 = 23 - 6x
6x = 0
X = 0
Ответ ( 0 ; 23 )