2Cos5x Cosx - Cos5x = 0
Cos5x(2Cosx -1) = 0
Cos5x = 0 или 2Cosx -1 = 0
5x = π/2 + πk , k ∈Z Cosx = 1/2
x = π/10 + πk/5, k ∈Z x = +-arcCos1/2 + 2πn , n ∈Z
x = +-π/3 + 2πn , n ∈Z
Ответ:........................
1
4x²-24xy+9y²-6x²+10xy-3xy+5y²=-2x²-17xy+14y²
2
xy²(25x²-4y²)=xy²(5x-2y)(5x+2y)
5(9-6a+a²)=5(3-a)²
3
4x²-9-4x²-8x-4+1=0
-8x=12
x=-1,5
4
{4x+y=-10/*2⇒8x+2y=20
{5x-2y=-19
прибавим
13x=39
x=3
12+y=-10
y=-22
(3;-22)