Need some help solving high-school level trignometry question. here it is. 

I've tried solving it multiple ways but it gets too complicated. Is there any way to solve this?
 A: $$\displaystyle \overline{AB}=2rcos(x)=R$$, where R is radius of the shaded circle.
We have $2R+R(2x)=\frac{2\pi r}{2}$ by the condition given, rearrange to get the equality you want to prove.
A: Let $U = 2 \pi * r$ be the circumference of the circle.
Let $s = AB = AC$ be the radius of the circular arc.
Let $V = \frac{U}{2}$ be the perimeter of the shaded area.


*

*Find an equation for $V$:


$$
V = AB + AC + (2 \pi * s) * \frac{2x}{2 \pi} = 2 * s + 2 * x * s = s * (2 + 2x)
$$
It is the sum of $AB$ and $AC$ (which are both $s$) and the partial circumference of the arc $(2 \pi * s) * \frac{2 x}{2 \pi}$.


*Find an equation for $s$


$$
s = 2 * \cos{x} * r
$$
$OCA$ is an equilateral triangle since both $A$ and $C$ lie on the circumference of the circle with center $O$ and radius $r$
So $\cos{x} * r$ is the length of the adjacent side of a right-angled triangle $ODA$ where $D$ lies in the center of $AC$.


*Equate both terms for $V$


$$
\frac {U}{2} = s * (2 + 2x) \\ \Leftrightarrow
\pi * r = s * (2 + 2x) \\ \Leftrightarrow
\pi * r = 2 * \cos{x} * r * (2 + 2x) \hspace{10pt} \vert * \frac{1}{r} \\ \Leftrightarrow
\pi = \cos{x} * (4 + 4x) \hspace{10pt} \vert * \frac{1}{4 + 4x} \\ \Leftrightarrow
\frac{\pi}{4 + 4x} = \cos{x} \hspace{10pt} \vert \arccos{} \\ \Leftrightarrow
\arccos{\frac{\pi}{4 + 4x}} = x
$$
