Critical numbers of the functionF(o) = 8 sec θ + 4 tan θ, 0 < θ <2 π.F(o) = 0 for critical numbersF(o) = 8 sec θ tan θ + 4 sec^2 θ = 0 = 4 sec θ [2 tan θ + 4 sec θ] = 0Sec θ = 0 or 2 tan θ + sec θ = 0Sec θ = 0 is no solution.2 tan θ + sec θ = 0 => (2 sin θ / cos θ) + (1 / cos θ) = 0 (2 sin θ + 1 / cos θ) = 0 => 2 sin θ + 1 = 0 Sin θ = -1/2 Θ = 7π/6 , 11π/6 Therefore, critical number are 7π/6 , 11π/6
