Simple trigonometric inequality, yet no simple solution.

Hello, Doubled144314!

I will limit my answers to Quadrant 1.
You can generalize the results if you like.

Quote:

$\cos x+\tan x \:>\:\sin x+1$

We note that $x \ne \tfrac{\pi}{2}$

We have: . $\cos x + \frac{\sin x}{\cos x} \:>\:\sin x + 1$

Multiply by $\cos x\!:\;\;\cos^2\!x + \sin x \:>\:\sin x\cos x + \cos x$

. . . . $\sin x - \sin x\cos x - \cos x + \cos^2x \:>\:0$

. . . . $\sin x(1-\cos x) - \cos x(1-\cos x) \:>\:0$

. . . . . . . . . . $(1-\cos x)(\sin x - \cos x) \:>\:0$

There are two cases: . $\begin{bmatrix}\text{(1) both factors are negative} \\ \text{(2) both factors are positive}\end{bmatrix}$

Case (1): . $1-\cos x \:<\:0 \quad\Rightarrow\quad \cos x \:>\:1\;\text{ impossible}$

Case (2): . $1 - \cos \:>\:0 \quad\Rightarrow\quad \cos x\:<\:1 \quad\Rightarrow\quad x \in \left(0,\,\tfrac{\pi}{2}\right]$

. . . . . . . . $\sin x-\cos x \:>\:0 \quad\Rightarrow\quad \sin x \:>\:\cos x \quad\Rightarrow\quad \frac{\sin x}{\cos x} \:>\:1$
. . . . . . . . $\tan x \:>\:1 \quad\Rightarrow\quad x \in \left(\tfrac{\pi}{4},\:\tfrac{\pi}{2}\right)$

Therefore: . $x \in \left(\tfrac{\pi}{4},\:\tfrac{\pi}{2}\right)$