This system is from the last century. This one has already appeared several times on the forum.
It is clear that nowadays it is not difficult to find several of its solutions, especially with the help of Maple. But in this case we are talking about a complete solution in real numbers. Of course, there are techniques that practically allow us to talk about such a solution, only without theoretical justification. 
The simple idea of ​​transforming it from transcendental to polynomial and getting a complete solution on a theoretical basis did not lead to a good result: it turned out that this is too complicated work for the  Isolate procedure. Perhaps I made a mistake somewhere, or (and) my old PC is too weak.
(The values ​​of the constants are not very important.)

restart; with(RootFinding):
 b1 := 114.069^2; b2 := 109.2389^2; b3 := 103.892^2; b4 := 99.76348^2; b5 := 97.24296^2; 
f1 := x1^2+x2^2+x3^2+2*(x1*x2*cos(x4)+x2*x3*cos(x5)+x1*x3*cos(x4+x5))-b1; 
f2 := x1^2+x2^2+x3^2+2*(x1*x2*cos(2*x4)+x2*x3*cos(2*x5)+x1*x3*cos(2*(x4+x5)))-b2; 
f3 := x1^2+x2^2+x3^2+2*(x1*x2*cos(3*x4)+x2*x3*cos(3*x5)+x1*x3*cos(3*(x4+x5)))-b3; 
f4 := x1^2+x2^2+x3^2+2*(x1*x2*cos(4*x4)+x2*x3*cos(4*x5)+x1*x3*cos(4*(x4+x5)))-b4; 
f5 := x1^2+x2^2+x3^2+2*(x1*x2*cos(5*x4)+x2*x3*cos(5*x5)+x1*x3*cos(5*(x4+x5)))-b5; 
f := seq(cat(f, i), i = 1 .. 5):
# fsolve([seq(f[i], i = 1 .. 5)]);
for i to 5 do 
F[i] := expand(f[i]);
F[i] := subs(cos(x4) = x4, cos(x5) = x5, F[i]); 
F[i] := subs(sin(x4) = sqrt(-x4^2+1), sin(x5) = sqrt(-x5^2+1), F[i]);
F[i] := subs(sqrt(-x4^2+1) = y4, sqrt(-x5^2+1) = y5, F[i]); 
F[i] := collect(F[i], [y4, y5]); 
F[i] := subs(y4 = sqrt(-x4^2+1), y5 = sqrt(-x5^2+1), F[i]); 
F[i] := op(1, F[i])^2-(sum(op(k, F[i]), k = 2 .. nops(F[i])))^2 
end do:
for i to 5 do 
F[i] 
end do;
#fsolve([seq(F[i], i = 1 .. 5)]);
#T := Isolate([seq(F[i], i = 1 .. 5)], [x1, x2, x3, x4, x5]): j := nops(T);
# T;

Edited: "+" to "-".

The problem arose while playing with inscribed circles, as in this post. If anyone is interested, try to find a circle of maximum radius inscribed between the curves
x1^2 + 2*x2^2 - 1 = 0 and (x1 - sin(x1))^2 + (x2 - sin(x2))^2 - 1 = 0.

Curve graphs.

A little continuation of topics 1 and 2. This is a very similar cube from 2, 
 

but with a different equation:

f1 := (x1-sin(x1))^2+(x2-sin(x2))^2+(x3-sin(x3))^2-0.02513144866;
And other point coordinates (-.8283302152, -.8283302152, .8283302152) and (.8283302152, .8283302152, -.8283302152).

I liked the recent question from user goebeld and especially the answer from Rouben Rostamian.
I admit, I didn’t even realize that Maple had VariationalCalculus procedures.
But what if the red and green  points are on the surface x1^4 + x2^4 + x3^4 -1 = 0 ? 
Points coordinates (-0.759835685700000, -0.759835685700000, 0.759835685700000) and
 (0.759835685700000, 0.759835685700000, -0.759835685700000).

Where will the shortest distance between these points on a given surface be? Taking into account symmetry, of course.

It was found on the social networks of the WM group. Written in Python. Perhaps someone would like to adopt it.