Short answer:
Unfortunately, there is no closed solution for the given problem!
A better precision of the solution is 0.7390851332151607.
However, if you need a much higher precision, you can work with Newton's method that allows you to calculate the value of the solution to any degree of accuracy you want, provided you have the tools (computer software) to work with the precision you need.
This is how it works.
f(x) = cos(x)-x = 0
f'(x) = -sin(x) -1
Given an initial approximation of solution x0=0.7, we calculate the next approximation using Newton's formula:
x1=x0-f(x0)/f'(x0)
In the present case, using a crude starting approximation of x=0.7, we can calculate successively better approximation using very few approximations, or iterations. Digits in bold indicate accurate digits.
x0=0.7
x1=0.7-f(0.7)/f'(0.7)=0.7394364978480581
x2=x1-f(x1)/f'(x1)=0.7390851604651073
Seeing that accuracy increases by doubling digits, we increase the calculation precision to 100 digits.
x3=0.739085133215160805616473437040986470634676756320239380701092376720213084353767950446652444225306274
From now on, we can only determine the number of accurate digits by comparing with the next iteratioin.
x4=0.739085133215160641655312087673879339982814654240052551704075312279704848098342239352991371439430969
x5=0.739085133215160641655312087673873404013411758900757464965680635781064894802035850700625476229759288
x6=0.739085133215160641655312087673873404013411758900757464965680635773284654883547594599376106931766532
x7=0.739085133215160641655312087673873404013411758900757464965680635773284654883547594599376106931766532
We see that both x6 and x7 are accurate up to 100 digits.
In fact, we can obtain up to 1000 digits of accuracy with just about 10 iterations. Even less iterations if we start with a better estimate instead of the crude one of 0.7.
