You have one equation with three unknowns. Such equations usually have infinitely many solutions. The plot clearly shows that your equation is satisfied by the points lying on  some surfaces:

plots:-implicitplot3d((x,y,z)->`if`(y>x and z>y,cos(2*Pi*(x+y-2*z))+cos(2*Pi*(y+z-2*x))+cos(2*Pi*(z+x-2*y)),undefined),  0..1, 0..1, 0..1, style=surface, grid=[50,50,50], axes=normal);

restart;
h:=k-(k-1)*x: lambda:=k/(1+k):
DE:=diff(p(x),x)=6/h^2-12/h^3*lambda;
BC:=p(0)=0, p(1)=0;
simplify(dsolve({DE, BC}, p(x)));
is(eval(p(x),%)=6*(h-1)*(h-k)/h^2/(1-k^2));

Or

restart;
h:=k-(k-1)*x: lambda:=k/(1+k):
DE:=diff(p(x),x)=6/h^2-12/h^3*lambda;
BC:=p(0)=0, p(1)=0;
simplify(dsolve({DE, BC}, p(x)));
p:=unapply(eval(p(x),%), x);
h:='h':
eval(p(x),x=solve(h=k-(k-1)*x,x));
simplify(%);

int(u*v, x=-1..0, y=-1..1) + int(u*v, x=0..1, y=-1..1);

expr:=sin(x)+cos(x)*exp(x)+x*tan(x)*f(x)+x+y;
indets(expr, function) union {x};

                         {x, cos(x), exp(x), f(x), sin(x), tan(x)}

restart;
plot3d([[r*cos(phi),r*sin(phi),r],[r*cos(phi),r*sin(phi),2-sqrt(4-r^2)]], r=0..2, phi=0..2*Pi, scaling=constrained);

Another option:

plot3d([[r*cos(phi),r*sin(phi),r],[r*cos(phi),r*sin(phi),2+sqrt(4-r^2)]], r=0..2, phi=0..2*Pi, scaling=constrained);

@flesx  R is your initial region.

restart;
R:=plot([seq(k,k=0..1,0.1),seq([k,y,y=0..1],k=0..1,0.1)],x=0..1, color=blue, scaling=constrained):
z:=x+I*y:
F:=plottools:-transform(unapply(evalc([(Re,Im)(exp(2*z))]), x,y)):
plots:-display(<R | F(R)>);

The code:
plots:-display(plottools:-curve([[x1,y1,z1], [x2,y2,z2], ... ,[xn,yn,zn]]));
 

[[x1,y1,z1], [x2,y2,z2], ... ,[xn,yn,zn]]  is the list of lists of your points.

As for the statement of the classical problem see wiki for details.

We assume that:  newly born  n  pairs of rabbits are put in a field; each  pair mates at the age of one month, and at the end of their second month they  produce another pair of rabbits with the probability 3/4 ; and rabbits never die, but continue breeding forever. So we come to the recurrent equation  u(k)=u(k-1)+3/4*u(k-2)  with the initial conditions  u(0)=n, u(1)=n

restart;
u:=unapply(rsolve({u(k)=u(k-1)+3/4*u(k-2),u(0)=n,u(1)=n}, u(k)),k);
expand(u(12));
round(%/n);

Thus, the original number of rabbits will increase approximately 97 times.

restart;
dat := <14.52407334|-162.1310124>:
A := 10^(dat[1]/20.):
phi := dat[2]*Pi/180.:
R:= 0.3036:
f := (Zeta,A,phi) -> cos(Zeta) - R * Zeta * sin(Zeta) - exp(-phi * I)/A:
soln := [solve(f(Zeta,A,phi), Zeta)]:
nos := nops(soln):
# soln := Vector(soln);
soln1:=select(s->Re(s)>0 and Im(s)>0, soln);
L:=sort(soln1,key=(s->Re(s)));
real_part:=Re(L[1]);
imaginary_part:=Im(L[1]);

We can numerically parameterize this curve by polar angle and then do any animations, for example:

restart; 
f1 := sqrt((x+0.500000000)^2+y^2)+sqrt((x-0.500000000)^2+(y-1)^2)+sqrt((x-1.050000000)^2+(y+1)^2)+sqrt((x-0.500000000)^2+(y+0.500000000)^2)+sqrt((x-1)^2+(y-1)^2)-7: 
X := t->fsolve(eval(f1, [x = r*cos(t), y = r*sin(t)]), r = 0 .. 20)*cos(t): 
Y := t->fsolve(eval(f1, [x = r*cos(t), y = r*sin(t)]), r = 0 .. 20)*sin(t): 
plots:-animate(plot, [[X, Y, 0 .. a]], a = 0 .. 2*Pi, frames = 60);

The code below works in Maple 2018.2:
 

restart;
F := proc(t)
local A, B, C;
uses plottools, plots;
A:=line([-2,0], [cos(t)-2, sin(t)], color=blue,thickness=3);
B:=line([cos(t)-2, sin(t)], [t, sin(t)], color=blue,thickness=3);
C:=plot(sin(x), x=0..t, view=[-3..7, -5..5],thickness=3);
display(A,B,C);
end proc:

plots:-animate(F,[theta],theta=0..2*Pi,background=plot([cos(t)-2,sin(t),t=0..2*Pi],thickness=3),
scaling=constrained,axes=none);

You can just skip this option:

restart;
p1 := plot([[0,0], [1,1]]);

If you want the plot of a function  f(x)  without labels, then define the function in operator form, for example:

restart;
plot(x->x^2, -2..2);

restart;
f := a*b+a*c+a+b+c;
s:=select(has, {op(f)}, a) minus {a*b};
# or
s:=select(has,{op(f)} minus {a*b}, a);

Output:       {a, a*c}

restart;
eq2 := ln(2*u^2 + u - 1) = -c - 2*ln(x);
sol:=[solve(eq2,u)];
simplify(sol[1]);
applyop(t->sqrt(simplify(t^2)), [2,2], %);

restart;
f:=x->3+(exp(x^2)+exp(1))/(exp(x^2)-exp(1));
P:=plot(f, -5..5, -3..8, color=blue, discont);
g:=x->f(a)-1/D(f)(a)*(x-a); # Equation of normal to f(x) at the point x=a
x0:=0:
M:=f(0);
d:=sqrt(a^2+(f(a)-g(0))^2):
a:=fsolve(g(0)-M=d, a=1..2);
R:=evalf(d);
y0:=evalf(g(0));
# (x-x0)^2+(y-y0)^2=R^2 - Equation of the circle
plots:-display(P, plottools:-circle([x0,y0],R, color=red), scaling=constrained);

We use a numerical solution ( fsolve  instead of  solve ) because the  solve  command cannot find any solution (returns NULL)