1) y-2x=6, x^2-xy+y^2=12;
2) y=6+2x, x^2-xy+y^2=12;
3) y=6+2x, x^2-x(6+2x)+(6+2x)^2=12;
4) y=6+2x, x^2-6x-2x^2+12+2x^2=12
5) y=6+2x, x^2-2x^2+2x^2-6x+12-12=0;
6) y=6+2x, x^2-6x=0;
7) y=6+2x, x(x-6)=0;
8) y=6+2x, x=0, x=6
9) y=8, y=18, x=0, x=6.
Решение
∧1 + (2*cos∧2x - 1) - (3/2) * cosx = 0
2 + 4*cos∧2x - 1 - 3*cosx = 0
2 + 4 *cos∧2x - 2 - 3*cosx = 0
cosx*(4cosx - 3) = 0
cosx = 0
x = π/2 + πn, n∈Z
π/2 ∈ [0,3;π/2]
cosx = 3/4
x = (+ -) arccos3/4 + 2πk, k∈Z не принадлежат промежутку:
Ответ: x = π/2 + πn, n∈Z
2х*х*4у*у^2*х^2=2*х^2*4*у^3*х^2=2*4*х^2*х^2*у^3=8х^4у^3