Find the domain of $f(x)$ $$f(x)=\sqrt{\cos(\sin x)}+\arcsin (\frac{1+x^2}{2x})$$

My Approach:

$\cos(\sin x) \geq 0 ; -1\leq \frac{1+x^2}{2x} \leq 1$

The first inequality holds for all $x$ and, for second one we have:

$(x+1)^2 \geq 0$ and, $(x-1)^2 \leq 0$ which is true only for $x=1$

My Problem:

The answer mentioned in the book is $x=1,-1$

I want to know how $-1$ is in the domain.

Find the domain of $f(x)$ $$f(x)=\sqrt{\cos(\sin x)}+\arcsin \left(\frac{1+x^2}{2x}\right)$$

My Approach:

$\cos(\sin x) \geq 0 ; -1\leq \frac{1+x^2}{2x} \leq 1$

The first inequality holds for all $x$ and, for second one we have:

$(x+1)^2 \geq 0$ and, $(x-1)^2 \leq 0$ which is true only for $x=1$

My Problem:

The answer mentioned in the book is $x=1,-1$

I want to know how $-1$ is in the domain.