<s>[INST] Solving a trig equation that is quadratic? I have to solve for $x$  given
$$\tan^2 x  = 2 + \tan x\;\;\;\;\;\;0≤x≤2\pi$$
I brought it all to one side and set it all equal to zero like:
$$\tan^2 x - \tan x - 2 = 0$$
What am i supposed to do from there to solve for x?
 [/INST] You began with:
$ \tan^2 x = 2 + \tan x $.

As you mentioned, bring all terms to the L.H.S.:
$ \tan^2 x - \tan x - 2 = 0 $

Now let:
$ y = \tan z $.

As a result:
$ y^2 - y - 2 = 0 $

Use the quadratic formula to solve for $y$.
$ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} $
$ = \frac{1 \pm \sqrt{9}}{2} $
$ = \frac{1 \pm 3}{2} $
$ = \frac{1 + 3}{2}, \frac{1 - 3}{2} $
$ = \frac{4}{2}, \frac{-2}{2} $
$ = 2, -1 $

Since $y = 2, -1 $, and $ y = \tan z $:
$ \tan z = 2, -1 $

From there:
$ z = \arctan 2, \arctan-1 $
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