Inequality: $\tan(x) > 1$

Question

So far, I've not come very... far. It ends up with me trying to solve it more intuitively than mathematically.

I figured, first I'll find the place of equality, which is at $x = \arctan 1 = \frac{\pi}{4} + \pi n$.

Then, it will be larger than 1 until either sin(x) or cos(x) changes sign, but I can't find a way to express this consistently over a larger interval.

Answer 1

Let us look in the interval $(-\pi/2,\pi/2)$, since $\tan$ is $\pi$-periodic. In this interval, if we want $\tan(x) > 1$, then $$x \in (\pi/4,\pi/2)$$ Hence, the desired intervals are $$n \pi + (\pi/4,\pi/2)$$

Answer 2

The principal domain of $y = \tan x$ is $(-\pi/2,\pi/2)$. Since $\tan (\pi/4) = 1$ and $y = \tan x$ is increasing, the solution to $\tan x > 1$ in this interval is $(\pi/4,\pi/2)$.

Since the tangent function is periodic with period $\pi$, the solution set in a different period has the form $(\pi/4+k\pi,\pi/2+k\pi)$. Thus $$ \{x : \tan x > 1\} = \bigcup_{k \in \mathbb Z} \left( \frac{\pi}{4} + k\pi, \frac{\pi}{2} + k\pi \right).$$

Answer 3

$\tan x$ is increasing in $[0,\pi/2)$ and principal solution of $\tan x>1$ is $[pi/4,pi/2)$ and we know the graph of $\tan x$ is repeated after every $\pi$ unit so final solution is $[n\pi+\pi/4,n\pi+\pi/2)$