Дана функция f (x) = 2/3cos (3x-П/6). Составьте уравнение касательной к графику функции в точке с абсциссой x=П/3

Y (x) = 2 / (3cos (3x - π/6))

ОДЗ: cos (3x - π/6) ≠ 0, 3x - π/6 ≠ π/2 + πk

3x ≠ π/2 + π/6 + πk, 3x ≠ 2π/3 + πk, k∈Z

x ≠ 2π/9 + πk/3, k∈Z

Y = y (a) + y' (a) * (x - a) - уравнение касательной

a = π/3

y (π/3) = 2 / (3*cos (3*π/3 - π/6)) = 2 / (3*cos (π - π/6)) = - 2 / (3*cos (π/6)) = - 2*2 / (3*√3) = - 4√3/9

y' (x) = (2/3) * (-1) * (cos^ (-2) (3x - π/6)) * 3 = - 2/cos^2 (3x - π/6)

y' (π/3) = - 2/cos^2 (3*π/3 - π/6) = - 2/cos^2 (π/6) = - 2*4/3 = - 8/3

Y = - 4√3/9 - (8/3) * (x - π/3) = - (8/3) * x - (4√3/9) + (8π/9)