$\displaystyle sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1 $Solve the following equations in the given intervals.

$\displaystyle sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1 $

$\displaystyle 0\leq \theta \leq 2\pi $

$\displaystyle 1 + tan^2 \theta = sec^2 \theta $

$\displaystyle 1 + tan^2 \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1 $

$\displaystyle tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0 $

$\displaystyle tan^2 \theta - tan \theta - \sqrt{3}tan \theta = -\sqrt{3} $

$\displaystyle tan \theta ( tan \theta-\sqrt{3} -1 ) = -\sqrt{3} $

so

$\displaystyle tan \theta = -\sqrt{3} $

or

$\displaystyle tan \theta = 1 $

is this method correct? thanks!