1) y' = √3/2*1/Cos²x -2√3/3
2) √3/2*1/Cos²x -2√3/3 = 0
√3/2*1/Cos²x = 2√3/3
Cos²x = 3/4
Cosx = +-√3/2
3) a) Cosx = √3/2 б) Cosx = -√3/2
x = +-π/6 + 2πk , k ∈Z x = +-5π/6 + 2πn , n ∈Z
Из всех решений в указанный промежуток попал х = π/6
4) а) х = 0
у = √3/2*tg0 -2√3/3*0 +√3/9*π +1 = √3/9*π +1
б) х = 3π/4
у = √3/2*tg3π/4 -2√3/3*3π/4 +√3/9*π +1 = -√3/2*1 -2√3/3*3π/4 +√3/9*π +1
в) х = π/6
у =√3/2*tgπ/6-2√3/3*π/6 +√3/9*π +1 =√3/2*1/√3 +1 = 1,5
5) Ответ: min y = 1,5