Solving $\tan x> -\sqrt 3$

Question

$\tan x> -\sqrt 3$

How do I solve this inequality?

From the graph it is evident that $\tan x>-\sqrt 3$ for $\left(\dfrac{2\pi}3 , \dfrac{5\pi} 3\right)$ $\forall x\in (0, 2\pi)$.

Generalising this solution we get $\left(2n\pi +\dfrac{2\pi}3 , 2n\pi+\dfrac{5\pi} 3\right) \forall n \in \mathbb{Z}$ as the answer.

But the answer given is: $\left(n\pi - \dfrac \pi 3, n\pi + \dfrac \pi 2\right)$

Where have I gone wrong?

Answer 1

Tangent has period $\pi$, not $2\pi$. Also, $tan(x) > -\sqrt{3}$ when $x>-\pi/3$. The function is not defined at $\pi/2$, but it’s positive for $x \in [0, \pi/2)$.

Answer 2

You went wrong in reading the graph. The number $5\pi/3$ should have been $5\pi/2$. And the period is $\pi $ and not $2\pi $, which leads to the correct answer that you quoted: if you have $$\left( n\pi+\frac {5\pi}3,n\pi+\frac {5\pi}2\right), $$ Now,replacing $n $ with $n-2$ achieves the solution given.

Answer 3

First a side note: when you are working with graphs like this you should always work with the closest points to $0$, it will make your life easier

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So $\tan x>-\sqrt 3$, to find the values of $x$ first let's find where those 2 are equal, you will get $-\frac{\pi}3$(this is the closest point to $0$, ofc there are infinity many points like this).

Now $tan x$ is increasing function hence we found the lower bound, now we search for the first point where $tan x<-\sqrt 3$ and $x>-\frac{\pi}3$, but if $\tan x$ is strictly increasing we need to check only after a point it doesn't exists, the first point that doesn't exist and $x>-\frac{\pi}3$ is $\frac{\pi}2$, where after that point we are at $-\infty$, so this point have the 2 conditions that we look for. Hence we have the solution $(-\frac{\pi}3,\frac{\pi}2)$. Now $\tan x$ is periodic for $\pi$ so we get: $$x\in(-\frac{\pi}3+n\pi,\frac{\pi}2+n\pi)$$

Answer 4

There is an error when analyzing the function at $\pi/2$, since at that point there is a discontinuity. The correct way to proceed is analyzing the graph between $-\pi/2$ and $\pi/2$, in this way you get that the solution of $\tan(x)>-\sqrt{3}$ is $(-\pi/3,\pi/2)$ and adding the period $\pi$ you get $(n\pi-\pi/3,n\pi+\pi/2)$ $\forall n \in \mathbb{Z}$.