I was asked to find the domain of $$\arcsin[\frac{x^2+1}{2x}]$$ My first step was $$-1\leq\frac{x^2+1}{2x} \leq1$$
What I don't understand is why I cannot cross multiply to get $$-2x\le {1+x}^{2} \le 2x$$ and then solve the inequality? I tried doing this and got the wrong answer.
In order to find the domain, you need to check every possible real value of $x$. This includes values of $x$ where $x\ge 0$ and where $x<0$. When you multiplied all three terms of
$$-1\leq\frac{x^2+1}{2x} \leq1$$
by $2x$ to form
$$-2x\le {1+x}^{2} \le 2x$$
you only considered the case where $x\ge 0$. Since an inequality flips sign when multiplied by a negative number, multiplying all of the terms of
$$-1\leq\frac{x^2+1}{2x} \leq1$$
by $2x$ when $x<0$ forms
$$-2x\ge {1+x}^{2} \ge 2x$$
Therefore, you need to break up the analysis into two cases and consider what happens when $x\ge 0$ and $x<0$.
Case 1: Suppose $x\ge 0$. Then
$$-2x\le {1+x}^{2} \le 2x$$
by the second inequality
$${1+x}^{2} \le 2x$$
we see that $x=1$ is the only solution.
Case 2: Suppose $x < 0$. Then
$$-2x\ge {1+x}^{2} \ge 2x$$
where from the first inequality
$$1+x^2 \le -2x$$
we have that $x=-1$ is the only solution.
So, the domain of $\arcsin\left(\frac{x^2+1}{2x}\right)$ is $x\in\{-1,1\}$.
Let $x=\tan y$
$z=\arcsin\dfrac{1+x^2}{2x}=\arcsin(\csc2y)$
$\implies\sin z=\csc2y$ which is either $\ge1$ or $\le-1$
But $-1\le\sin z\le1$
So, $\dfrac{1+x^2}{2x}=\pm1$
Let $y=\arcsin\dfrac{x^2+1}{2x}$
$x^2-2x\sin y+1=0$
$x=\sin y\pm i\cos y$
As $x$ is real, $\cos y$ must be $0$
$\implies\sin y=\pm1$
