Solve the following equation over the set of real numbers:

$$
x-\frac{x}{\sqrt{x^{2}-1}}=\frac{91}{60}
$$
Let $u=\sqrt{x^{2}-1}$. Squaring both sides and substituting, we rearrange the equation as follows:

$$
x^{2}\left(1-\frac{1}{\sqrt{x^{2}-1}}\right)^{2}=\left(\frac{91}{60}\right)^{2} ;\left(u^{2}+1\right)\left(1-\frac{1}{u}\right)^{2}=\left(\frac{91}{60}\right)^{2} ;\left(u+\frac{1}{u}\right)\left(u+\frac{1}{u}-2\right)=\left(\frac{91}{60}\right)^{2}
$$

This is a quadratic equation in $\left(u+\frac{1}{u}\right)$, whose roots are $\frac{169}{60}$ and $-\frac{49}{60}$. However, by definition, $u$ and $u+\frac{1}{u}$ are positive, so the negative root cannot be considered. Therefore,

$$
u+\frac{1}{u}=\frac{169}{60}, \quad \text { i.e. } \quad u^{2}-\frac{169}{60} u+1=0
$$

This quadratic equation has two roots: $\frac{5}{12}$ and $\frac{12}{5}$. 

If $u=\frac{5}{12}$, then $x= \pm \sqrt{u^{2}+1}= \pm \frac{13}{12}$. By substitution, it can be verified that $-\frac{13}{12}$ is a solution, while $+\frac{13}{12}$ is not. If $u=\frac{12}{5}$, then $x= \pm \sqrt{u^{2}+1}= \pm \frac{13}{5}$. Among these, $+\frac{13}{5}$ is a solution, while $-\frac{13}{5}$ is not.

Therefore, the equation has two roots: $x_{1}=-\frac{13}{12}$ and $x_{2}=\frac{13}{5}$.