Dtjytyk

I have the following equation and I want to solve for $R$.

$$R^{2} left( frac{arccos left(frac{h}{R}right) + frac{h}{R}}{2} right) = pi r^{2}$$

As said in comments, this is a transcendental equation and you cannot expect analytical solutions and then some numerical method would be required.

We can make the equation looking nicer using $R=frac h x$ and $a=frac{pi r^2}{h^2} $which makes
$$a x^2-x-cos ^{-1}(x)=0$$ and the solution is the intersetion of $f(x)=xcos ^{-1}(x)$ with $g(x)=ax^2$. There is only one solution to the equation provided $a geq 1$ and, for sure, the solution is in $[0,1]$.

For sure, if $a$ is large (that is to say $x$ small), you could use Taylor expansion to get
$$a x^2-x-cos ^{-1}(x)=frac{pi }{2}-a x^2-frac{x^3}{6}+Oleft(x^5right)$$ and ignoring the higher order terms solve either the quadratic equation giving as approximation $x=sqrt{frac pi {2a}}$ or even the cubic equation (but this would not be very pleasant).

So, for $ageq 1$, use Newton method with $x_0=sqrt{frac pi {2a}}$ for finding the zero of
$$f(x)=a x^2-x-cos ^{-1}(x)$$
$$f'(x)=2 a x+frac{1}{sqrt{1-x^2}}-1$$
$$x_{n+1}=x_n-frac{f(x_n)}{f'(x_n)}$$

Trying for $a=2$, this would give the following iterates
$$left(
begin{array}{cc}
n & x_n \
0 & 0.88622693 \
1 & 0.84308752 \
2 & 0.84055714 \
3 & 0.84055001
end{array}
right)$$

We could have a better approximation of the guess building the $[2,k]$ Padé approximant of the function and solve the resulting quadratic. Using $k=2$, this would give as an estimate
$$x_0=frac{3 a left( sqrt{3pi(96 a^3-pi)}-piright)}{72 a^3-pi}$$ For $a=2$, this would give $0.856358$.

Another thing which could be done is to use series reversion and get
$$x=t-frac{t^4}{6 pi }-frac{3 t^6}{40 pi }+frac{5 t^7}{72 pi ^2}-frac{5 t^8}{112
pi }+frac{7 t^9}{80 pi ^2}-frac{left(128+105 pi ^2right) t^{10}}{3456 pi
^3}+frac{2067 t^{11}}{22400 pi ^2}-frac{left(704+189 pi ^2right)
t^{12}}{8448 pi ^3}+frac{11 left(98+407 pi ^2right) t^{13}}{48384 pi
^4}+Oleft(t^{14}right)$$ where $t=sqrt{frac pi {2a}}$ Applied again for $a=2$, this would give $0.841464$. Using this, I am sure that Newton method would converge in very few iterations.

In order to check the quality of the approximation, let us give $x$ a value; from $x$, compute $a=frac{x+cos ^{-1}(x) } {x^2}$ then $t=sqrt{frac pi {2a}}$ and recompute $x$ according to the last expansion. Below are rproduced some values
$$left(
begin{array}{ccc}
x_{text{given}} & a & x_{text{calculated}} \
0.05 & 628.310 & 0.0500000000 \
0.10 & 157.063 & 0.1000000000 \
0.15 & 69.7879 & 0.1500000000 \
0.20 & 39.2360 & 0.2000000000 \
0.25 & 25.0899 & 0.2500000000 \
0.30 & 17.4012 & 0.3000000004 \
0.35 & 12.7610 & 0.3500000031 \
0.40 & 9.74550 & 0.4000000194 \
0.45 & 7.67423 & 0.4500000998 \
0.50 & 6.18879 & 0.5000004337 \
0.55 & 5.08573 & 0.5500016534 \
0.60 & 4.24249 & 0.6000056758 \
0.65 & 3.58157 & 0.6500179206 \
0.70 & 3.05183 & 0.7000529960 \
0.75 & 2.61819 & 0.7501492859 \
0.80 & 2.25547 & 0.8004077547 \
0.85 & 1.94438 & 0.8511043677 \
0.90 & 1.66793 & 0.9030757253 \
0.95 & 1.40450 & 0.9596060077
end{array}
right)$$

A problem remains when $a$ is close to $1$. For that case, we can perform the Taylor expansion around $x=1$ to get
$$a=1+sqrt{2} sqrt{1-x}+(1-x)+Oleft((x-1)^{3/2}right)implies x=1-a+sqrt{2 a-1}$$

Edit

Since I suppose that you would like accurate results, you will be using Newton method.

Using a quick and dirty non linear regression $(R^2=0.999997)$, you could use for the wole range
$$x_0=left(sqrt{frac pi {2a}} right)-frac{117}{1426}left(sqrt{frac pi {2a}} right)^5$$