find the critical point of this function f(Θ)=2secΘ+tanΘ 0<Θ<2π

Answer 1

f'(θ) = 2secθtanθ+sec²θ

secθ(2tanθ)+secθ) = 0

secθ = 0 DNE

tanθ=-secθ/2

sinθ/cosθ = -1/(2cosθ)

sinθ = -1/2

θ=7π/6. 11π/6

f(7π/6) = 2(-2/√3) + (-√3) = -1.732

f(11π/6) = 2(2/√3) + (-√3) = 1.732

Critical points: (7π/6, -1.732) and (11π/6, 1.732)