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.

Elaborating on player3236's comment: Let $u = \tan x$. Then your equation is $\sqrt{3} u^2 = 2u + \sqrt{3}$. Solve this quadratic equation, and then use $u = \tan x$ to solve for $x$.
