Answer
$\frac{11\pi}{6}$
Work Step by Step
1. $tan\theta+\sqrt 3=sec\theta \Longrightarrow tan^2\theta+2\sqrt 3tan\theta+3=sec^2\theta \Longrightarrow tan^2\theta+2\sqrt 3tan\theta+3=1+tan^2\theta \Longrightarrow tan\theta=-\frac{\sqrt 3}{3}$
2. $\theta=k\pi+\frac{5\pi}{6}$
3. Within $[0,2\pi)$, we have $\theta=\frac{5\pi}{6},\frac{11\pi}{6}$, check only $\theta=\frac{11\pi}{6}$ fits the original equation.
