Подставим координаты точки в первое, затем во второе уравнение
-6=k/-1
k=6
-6=6*(-1)+m
-6=-6+m
m=0
(9 - 3√x+3√x)/(3-√x)*(3 -√x)²=9(3-√x) = 9 (3 -sqrt(4/81))=9(3-2/9)=9*3 - 2=25.
2sin²x + (2 - √2)cosx + √2 - 2 = 0
2 - 2cos²x + (2 - √2)cosx + √2 - 2 = 0
2cos²x + (√2 - 2)cosx - √2 = 0
cosx = (2 - √2 ± √(2 - 4√2 + 4 + 8√2))/4 = (2 - √2 ± √(√2 + 2)²)/4 = (2 - √2 ± (√2 + 2))/4 = {1; -√2/2}
cosx = 1 => x = 2πn, n ∈ ℤ
cosx = -√2/2 => x = π ± π/4 + 2πk, k ∈ ℤ
Ответ: x = 2πn, n ∈ ℤ; x = π ± π/4 + 2πk, k ∈ ℤ