Решение 1. Область определения y = 2cos(x-п/3) D(y) = R 2. Область значения - 1 ≤ 2cos(x-п/3) ≤ 1 - 1/2 ≤ cos(x-п/3) ≤ 1/2 1) cos(x-п/3) ≥ - 1/2 - arccos(-1/2) + 2πk ≤ x - п/3 ≤ arccos(-1/2) + 2πk, k ∈ Z - 2π/3 + 2πk ≤ x - п/3 ≤ 2π/3 + 2πk, k ∈ Z - 2π/3 + π/3 + 2πk ≤ x ≤ 2π/3 + π/3 + 2πk, k ∈ Z - π/3 + 2πk ≤ x ≤ π + 2πk, k ∈ Z 2) cos(x-п/3) ≤ - 1/2 arccos(-1/2) + 2πk ≤ x - п/3 ≤ 2π - arccos(-1/2) + 2πk, k ∈ Z 2π/3 + 2πk ≤ x - п/3 ≤ 2π - 2π/3 + 2πk, k ∈ Z 2π/3 + 2πk ≤ x - п/3 ≤ 4π/3 + 2πk, k ∈ Z 2π/3 + π/3 + 2πk ≤ x ≤ 4π/3 + π/3 + 2πk, k ∈ Z π + 2πk ≤ x ≤ 5π/3 + 2πk, k ∈ Z
{х-5у=8
1) 7x+y=20
y=20-7x
2)x-5(20-7x)=8
x-100+35x=8
36x=108
x=3
3)y=20-7*3=20-21=-1
ответ: x=3; y=-1
{5х-8у=1
{х+2у=4
1)x+2y=4
x=4-2y
2)5(4-2y)-8y=1
20-10y-8y=1
-18y=-19
y=1 1/18
3)x=4-2*1 1/18=1 8/9