## Precalculus (6th Edition) Blitzer

$x=\pi$
Step 1. Using the identity $sec^2x=1+tan^2x$, rewrite the equation as $sec(x)=tan(x)-1$ and take the square on both sides to get $sec^2x=1+tan^2x-2tan(x)$; thus we have $tan(x)=0$ Step 2. For $tan(x)=0$, we can find all x-values in $[0,2\pi)$ as $x=0,\pi$ Step 3. Checking the above with the original equation, we can find $x=\pi$ as the only answer.