(UFGO) trigonometry The set of all values ​​of $x$, so that the angle
$\left(\frac{x}{x^{2}+1}\right)\pi\,$ rad belongs to the $1^\mathrm{st}$ quadrant, is represented by what interval?
This is an inequality that I did for 
$\frac{x}{x^{2}+1}π≤\frac{π}{2}$
$\frac{x}{x^{2}+1}-\frac{1}{2}≤0$  then $\frac{-x^{2}+2x-1}{2x^{2}+1}≤0$ and what now?
Answer: $\;\left] 0;1\right[ \cup \left]1;+\infty\right[$
 A: Angles in the first quadrant are by definition given by 
$$ \lbrace \theta\in\mathbb{R}\; :\; 0< \theta< \pi/2\rbrace.$$
To get the answer you want, angles $0$ and $\pi/2$ must be excluded from this set. 
By imposing this condition on your angle $\theta(x)=\left(\tfrac{x}{x^2+1}\right)π$, it follows that you have the inequality
$$0 <\left(\dfrac{x}{x^2+1}\right)π< \pi/2 \qquad\Leftrightarrow\qquad 0< \dfrac{x}{x^2+1}<1/2.$$
A little bit more elementary algebra shows that you want 
$$0< 2x<x^2+1$$
for all $x$. The first inequality means $x>0$ and from the second one, you have $x^2-2x+1>0$, which can be rewritten $(x-1)^2>0$. This is always the case except if $x=1$. You then have the common set given by the intersection: 
$$ \mathbb{R}_{>0}\cap (\mathbb{R}\setminus \lbrace 1\rbrace )=\rbrack 0,1\lbrack\cup\rbrack 1,\infty\lbrack$$
as desired.
A: HINT Angle $\theta$ belongs to the first quadrant is the same as $\,0\le \theta\le \dfrac{\pi}{2},$ i.e.
$$
0\le\left(\frac{x}{x^{2}+1}\right)\pi\le\frac{\pi}{2} \quad
\implies
\quad 0\le\frac{x}{x^{2}+1}\le\frac{1}{2} \quad
\iff\quad
\left\lbrace\begin{align}
\frac{x}{x^{2}+1} &\ge 0, &\text{and} \\
\frac{x}{x^{2}+1}&\le \frac{1}{2}.
\end{align}\right.
$$
Can you find all values of $x$ which satsfy two inequalities above?
