y=sin²x-cosx
y¹(x)=2sinx cosx+sinx=sinx(2cosx+1)=0
a) sinx=0, x=πn, n∈Z
b) 2cosx+1=0, cosx=-1/2, x=±(π-arccos1/2)+2πk, x=±(π-π/3)+2πk, x=±2π/3+2πk, k∈Z
Критические точки: х=πn, n∈Z , x=±2π/3+2πk, k∈Z.
1)3xy + 6ay = 3y(x + 2a)
2) y³ - y⁴ = y³(1 - y)
Решение
(π/3 - 0)^2 - 1 = (π)^2 / 4 - 1
Ax²+bx+c=0
x₁+x₂=-b/a
x₁x₂=c/a
5x²-2x-1=0
x₁+x₂=2/5
x₁x₂=-1/5
(x₁+x₂)²=x₁²+x₂²+2x₁x₂=(2/5)²
x₁²+x₂²-2/5=4/25
x₁²+x₂²=14/25
!x₁-x₂!=√(x₁-x₂)²=√(x₁²+x₂²-2x₁x₂)=√(14/25+2/5)=√(24/25)=√24/5