Question

помогите пожалуйста
)))

Answer 1

y'=(x³-3x²+2x-1)'=3x²-6x+2
y'(0)=3*0-6*0+2=2

2cos(π/3-3x)-√3=0
2cos(π/3-3x)=√3
cos(π/3-3x)=√3/2
π/3-3x=+-π/6+2πn, n∈Z
-3x=+-π/6-π/3+2πn, n∈Z
x=+-π/6+π/3+2πn, n∈Z
x₁=π/2+2πn, n∈Z
x₂=π/6+2πn, n∈Z

tg(x/4)+√3=0
tg(x/4)=-√3
x/3=2π/3
x=2π