Uncertain how to solve this trigonometric equation I am currently attempting to discover how to find the general solutions to $$\sqrt 3\tan^2x=2\tan x+\sqrt 3$$
The given solutions are $x= \frac{\pi}{3}+ \pi k$ , $\frac{5\pi}{6} + \pi k$
To solve this equation I removed the square root from $\sqrt 3\tan^2x$ leaving me with $\tan x$.
Then I subtracted $2\tan x$ from $3\tan x$ leaving me with $\tan x=\sqrt 3$.
Should I then not solve $\tan x = \sqrt 3$ for the generalized solutions?
This would give $x=\frac{\pi}{3}$ and $x=\frac{2\pi}{3}$ which would then be generalized to $\frac{\pi}{3} + \pi k$ and $\frac{2\pi}{3} + \pi k$, which doesn't agree with the given answers.

$$\text{Let }u=\tan x. $$
Divide through by $\sqrt{3}.$
$$ u^2 -\frac{2}{\sqrt{3}}u-1 = 0.$$
This is a quadratic equation, the solutions are:
$$u =  \frac{1}{\sqrt{3}} \mp \frac{2}{\sqrt{3}}.$$
We thus have$$ \tan x = \sqrt{3} \quad \text{ or } \quad \tan x = - \frac{1}{\sqrt{3}}.$$
So $$x \in \left\{\frac{\pi}{3} + k\pi, \frac{5\pi}{6} + k\pi\right\}.$$
