restart; 
simplify(A*sin(x)^2+A*cos(x)^2);

A simple procedure  Cone  returns parametric equations and plots a conical surface defined its vertix  S  and 3 points  A, B, С of the base lying on a plane  xOy . If  S  is a number then the procedure treats it as the height of a right circular cone. If  S  is a list, then it is the coordinates of the vertex of the cone.

Download Cone.mw

Edit.

Changes in your graphs when changing the parameter  E  is convenient to observe with the help of animation. I took a range for the animation parameter  E = 0 .. 2  with the step  0.02 . See attached file.

captive_breeding_new.mw

The  isolve  command fails with this equation.

Rewrite the equation as  2^y - x^2 = 615 . It is easy to check by the last digit that  y  should be even that is  y=2*n . So we have  (2^n - x)*(2^n + x) = 615 . The number 615 can be factored into 2 integer factors in 3 ways for  x>0:  615 = 3*205 = 5*123 = 15*41 . 

Solving the corresponding 3 systems, we obtain the unique solution:

restart;
ifactor(615);
solve({2^n-x=5,2^n+x=123});
solve({2^n-x=3,2^n+x=205});
solve({2^n-x=15,2^n+x=41});

Use the double  seq  command for this. First, I used a set structure to remove duplicate elements and for automatic sorting, then converted it into a list:

convert({seq(seq(a^2+b^2, a=0..5), b=0..5)}, list);
select(type, %, odd);

Use the  rsolve  command with the  makeproc  option:

restart;
u:=rsolve({u(n + 1) = 3.5*u(n)*(1 - u(n)), u(0) = 0.4}, u(n), makeproc);
seq([k,u(k)], k=0..100);
plot([%], style=point);

A:=<a,b; c,d>;
B:=<e, f>;
<A | B>; 

I have not found another solution to the problem than using  InertForm  package:

restart;
with(InertForm):
plots:-textplot([1,1,Typeset(Parse("Pi/6"))]);

csgn(x^2) assuming real;

  Output:  1                       

f:=x->3*x^2+1:
Matrix([seq([x,f(x)], x=1..3.5,0.5)]);

Another way is to turn  x(t)  into a name using oblique quotes:

restart;
f:= 2*`x(t)`+5;
diff(f, `x(t)`);

You can simplify the equation  eq  significantly if you run

simplify(eq) assuming h>0;

@ms0439883 
Here is a simple recursive procedure that solves your problem for any  n  and  k . It does not use any packages, only commands from the Maple core.

Download proc.mw

Look at this post  https://www.mapleprimes.com/posts/200677-Partition-Of-An-Integer-With-Restrictions , which presents the  Partition  procedure. The procedure solves a variety of problems associated with partition of an integer, in particular partition with different restrictions.

After the loading  screen appears, open the task manager, then the details, then 2 times destroy the process whose name begins with  jogamp_...