## Calculus 10th Edition

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