Why are there two solutions rather than one for $\operatorname{sech}^{-1}x=\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)$? $$\operatorname{sech}^{-1}x=\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)$$
When I tried to solve $x=\operatorname{sech}^{-1}\frac{2}{3}$, I got $\operatorname{sech}^{-1}x=\ln\left(\frac{3+\sqrt{5}}{2}\right)$, however it seems there is another solution, $\operatorname{sech}^{-1}x=\ln\left(\frac{3-\sqrt{5}}{2}\right)$
Is it because for values of $0<x<1$, $\sqrt{1-x^2} <1$ thus the formula below is valid?:
$$\operatorname{sech}^{-1}x=\ln\left(\frac{1±\sqrt{1-x^2}}{x}\right)$$
Since initially when deriving $\operatorname{sech}^{-1}x$, the sign ± is converted to + so that in instances where $\sqrt{1-x^2} \geqslant1$, $1+\sqrt{1-x^2}$ will never be zero for the purpose of taking logs?
 A: Since $\cosh x$ is an even function, so is its reciprocal $\mathrm{sech}\,x$, in which case $x=\mathrm{sech}\,y$ has two solutions for the variable $y$ which are opposites of each other. You can verify this for your two formulas with $-\ln u=\ln u^{-1}$ and rationalizing the denominator. Also since $\mathrm{sech}\,x$ is is not one-to-one on all of $\mathbb{R}$, it does not have an inverse function unless we restrict to a smaller domain on which it is. Just as with e.g. $\cos x$ or $x^2$ we restrict to the interval $[0,\infty)$ and so when solving for the inverse we choose the positive value.
If you choose $1+\sqrt{1-x^2}$ the log is positive whereas if you choose $1-\sqrt{1-x^2}$ the log is negative, so the formula for the inverse function will use the $+$ sign.
A: $f(x)=\frac{e^x+e^{-x}}{2}$ increases on $[0,+\infty)$ and decreases on $(-\infty,0]$.
$g(x)=\ln\frac{x+\sqrt{x^2-1}}{2}$ is an inverse function for $f$, which defined on $[0,+\infty).$
Since $f$ increases, the equation $f(x)=g(x)$ has an unique root .
But on $(-\infty,0]$ it's not always so. 
