How to find the maximum value of $x\cos^{-1}\left(x\right)$? I wanted to find the maximum value of $x\cos^{-1}\left(x\right)$ through differentiation, but upon on differentiation I get $$f'(x) = \arccos\left(x\right) - \frac{x}{\sqrt{1-x^2}}$$  to find the maximum I equated the function to zero:    $$\arccos\left(x\right) - \frac{x}{\sqrt{1-x^2}} = 0$$
But I am unable to find the roots of this equation. Could someone show how to find the zeros for this equation?

\begin{align*}
f(x) &=xcos^{-1}(x) \\
f'(x) &=cos^{-1}(x) - \frac{x}{\sqrt{1-x^{2}}} \\
0 &=cos^{-1}(x) - \frac{x}{\sqrt{1-x^{2}}} \\
\frac{x}{\sqrt{1-x^{2}}} &=cos^{-1}(x) \\
cos(\frac{x}{\sqrt{1-x^{2}}}) &= x \\
\end{align*}
Now substitute $x = cos\alpha$:
\begin{align*}
cos(\frac{cos\alpha}{\sqrt{1-(cos\alpha)^{2}}}) &= cos\alpha \\
cos(\frac{cos\alpha}{sin\alpha}) &= cos\alpha \\
\frac{cos\alpha}{sin\alpha} &= \alpha \\
cot\alpha &= \alpha \\
\end{align*}
I will admit that I do not know how to analytically find the $\alpha$ that makes this true (I looked online and the only answers I found were using Newtons method or other ways of approximating the answer). Either way, the cosine of this $\alpha$ is the maximum point of $f(x)$.
