Solve $\tan^2(x) - (1+\sqrt3)\tan(x) + \sqrt3 < 0$

Question

I've tried solving this as a square inequality:

$\tan^2(x) - (1+\sqrt3)\tan(x) + \sqrt3 < 0$.

...but I get $D = 4 - 2\sqrt3$ and I don't think that's the right way to go about this.

Answer 1

Maybe it means $\tan^2x - (1+\sqrt3)\tan{x} + \sqrt3 < 0$?

If so, use $$4-2\sqrt3=3-2\sqrt3+1=(\sqrt3-1)^2.$$

Answer 2

How about changing the question?

$(\tan x-1)(\tan x -\sqrt{3}) \lt 0$

So the zeroes of that function are $\dfrac{\pi}{4}$ and $\dfrac{\pi}{3}$.

If you look at the values in brackets, both are negative for $x \lt \dfrac{\pi}{4}$, meaning the final value is positive. For $x \gt \dfrac{\pi}{3}$ both values in brackets are positive, meaning the final value is also positive.

Let's analyse that interval $x \in (\dfrac{\pi}{4}, \dfrac{\pi}{3})$. In this interval, $(\tan x -1)$ is positive but $(\tan x-\sqrt{3})$ is negative, meaning the final result is negative and hence $\lt 0$

So here's your final answer:

$\dfrac{\pi}{4}+2\pi n \lt x \lt \dfrac{\pi}{3}+2\pi n \quad n \in \mathbb{N}$