#### Answer

$ θ=\frac{11\pi}{6} $, $ \frac{7\pi}{6} $

#### Work Step by Step

The first step is to take the derivative of the function $ f(θ)= 2sec(θ) + tan(θ) $.
Using the Basic Differentiation Rules you get $ f'(θ)=2sec(θ)tan(θ) + (sec( θ ))^{2} $
Next Simplify
$f'( θ )=2\frac{1}{cos( θ)} \times \frac{sin( θ )}{cos( θ )} + \frac{1}{(cos θ )^2} $
$f'( θ )= \frac{2sin(θ)}{cos^2(θ)} + \frac{1}{cos^2(θ)}$
$ f'(θ)= \frac{2sin(θ) +1}{cos^2(θ)} $
Then set the derivative equal to zero and solve for θ.
$0=\frac{2sin(θ)+1}{cos^2(θ)}$ , Multiply both sides by $cos^2(θ)$
$0=2sin(θ)+1$, Isolate the $sin(θ) $ term
$sin(θ)=-\frac{1}{2} $, take the $sin^{-1}(θ) $
of both sides to get
$ θ=\frac{11\pi}{6} $, $ \frac{7\pi}{6} $