Fractional mod!

$\left| \dfrac{1}{|x|-1} \right| = x + \sin x \quad \Rightarrow \left| \dfrac{1}{|x|-1} \right| - x = \sin x \quad \Rightarrow -1 \le \left| \dfrac{1}{|x|-1} \right| - x \le 1$

Let $f(x) = \left| \dfrac{1}{|x|-1} \right| - x$ and consider it for $x < -1, \quad -1 \le x < 0, \quad 0 \le x < 1$ and $x \ge 1$ .

$f(x) = \begin{cases} \dfrac{1}{-x-1} - x & \in [3,\infty) & \text{for } x <-1 \\ \dfrac{1}{x+1} - x & \in (1,\infty) & \text{for } -1 \le x <0 \\ \dfrac{1}{1-x} - x & \in [1,\infty) & \text{for } 0 \le x < 1 \\ \dfrac{1}{x-1} - x & \in (-\infty,\infty) & \text{for } x \ge 1 \end{cases}$

$\Rightarrow -1 \le \left| \dfrac{1}{|x|-1} \right| - x \le 1$ when $x \ge 1$ and actually when $\sqrt{2} \le x \le 2$ , and we have:

\(\begin{array} {} \text{When } x = \sqrt{2}, & f(\sqrt{2}) = 1 > \sin \sqrt{2} = 0.9878 \\ \text{When } x = 2, & f(2) = -1 < \sin 2 = 0.9093 \end{array} \)

Since $2 - \sqrt{2} = 0.1865 \pi$ , there is only $\boxed{1}$ value of $x$ that satisfies the equation.