What is the solution to the mentioned equation for $x$, given that $a > 0$ and $\frac{a}{2} < x < a$ ?
This equation arises from an effort to maximize the perimeter of a circular arc, as illustrated in the figure, which is tangent to a rectangular strip at one side, and whose center (and subsequently the value of radius $x$) varies vertically.
I also want the center to fall inside the strip. Also a circle with radius less than $\frac{a}{2}$ obviously results in a circle with less perimeter than one with radius $\frac{a}{2}$. So I simply assume that $\frac{a}{2} \leq x \leq a$.
According to my calculations, since the arc perimeter, $p$, is equal to $x\theta$, and $\theta = 2\arccos\left(\frac{x-a}{x}\right)$ ($-1 \leq \frac{x-a}{x} \leq 0$), we can obtain perimeter $p$ as a function of $x$ as follows $$p(x) = 2x\arccos\left(\frac{x-a}{x}\right).$$ Therefore $$p'(x) = 2\arccos\left(\frac{x-a}{x}\right) - \frac{2a}{\sqrt{2ax - a^2}}.$$ Since $p$ is continuous on $\left[\frac{a}{2}, a\right]$ and differentiable on $\left(\frac{a}{2}, a\right)$, and also $p(\frac{a}{2}) = p(a) = \pi a$, there is at least on point $x_0$ in interval $\left(\frac{a}{2}, a\right)$ such that $p'(x_0) = 0$. Therefore the equation of the question has solution.
My problem: I've got no idea how to solve the mentioned equation. Please help me. Thanks.
If you set $$ t=\frac{x-a}{x} $$ the equation becomes $$ \arccos t-\sqrt{\frac{1-t}{1+t}}=0 $$ for $-1<t<0$.
The function on the left-hand side can be seen to be decreasing over $(-1,0)$. Wolframalpha says that the root is, approximately, $$ t ≈ -0.689157736645164... $$ Use that $$ x=\frac{a}{1-t} $$