Answer
$\frac{\pi}{6}$
Work Step by Step
$\tan x+\sec x=\sqrt{3}$
$\tan x=\sqrt{3}-\sec x$
Square both sides:
$\tan^2 x=(\sqrt{3}-\sec x)^2$
$\tan^2 x=3-2\sqrt{3}\sec x+\sec^2 x$
$0=3-2\sqrt{3}\sec x+\sec^2 x-\tan^2 x$
$0=3-2\sqrt{3}\sec x+1$
$0=4-2\sqrt{3}\sec x$
$2\sqrt{3}\sec x=4$
$\sec x=\frac{2}{\sqrt{3}}$
$\cos x=\frac{\sqrt{3}}{2}$
The only solutions in $[0, 2\pi)$ are $\frac{\pi}{6}$ and $\frac{11\pi}{6}$.
Since squaring both sides is an operation that may introduce extraneous solutions, we need to check our answers by plugging in these two values into the original equation:
$LHS=\tan \frac{\pi}{6}+\sec \frac{\pi}{6}=\frac{\sqrt{3}}{3}+\frac{2\sqrt{3}}{3}=\frac{3\sqrt{3}}{3}=\sqrt{3}=RHS$
$LHS=\tan \frac{11\pi}{6}+\sec \frac{11\pi}{6}=-\frac{\sqrt{3}}{3}+\frac{2\sqrt{3}}{3}=\frac{\sqrt{3}}{3}\ne RHS$
So we reject $\frac{11\pi}{6}$ and the only solution in $[0, 2\pi)$ is $\frac{\pi}{6}$.
