Trig equation, is this correct?
Quote:
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 sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1 $
$\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!