@Markiyan Hirnyk 

Thank you!
You are right.  I did not notice that in the third quarter the plots are superimposed on one another. But Maple is also wrong, the solutions  {a=0, b=0}  and  {a=1, b=1}  are lost.

@Markiyan Hirnyk 

Thank you!
You are right.  I did not notice that in the third quarter the plots are superimposed on one another. But Maple is also wrong, the solutions  {a=0, b=0}  and  {a=1, b=1}  are lost.

@Markiyan Hirnyk

Try  RealDomain:-solve(sys, symbolic=false);

@Markiyan Hirnyk

Try  RealDomain:-solve(sys, symbolic=false);

@Markiyan Hirnyk 

Take a closer look:

sys := [sqrt(sin(x)^2+1/sin(x)^2)+sqrt(cos(y)^2+1/cos(y)^2) = sqrt(20*y/(x+y)), sqrt(sin(y)^2+1/sin(y)^2)+sqrt(cos(x)^2+1/cos(x)^2) = sqrt(20*x/(x+y))]:

A := lhs(sys[1])+lhs(sys[2]);

@Markiyan Hirnyk 

Take a closer look:

sys := [sqrt(sin(x)^2+1/sin(x)^2)+sqrt(cos(y)^2+1/cos(y)^2) = sqrt(20*y/(x+y)), sqrt(sin(y)^2+1/sin(y)^2)+sqrt(cos(x)^2+1/cos(x)^2) = sqrt(20*x/(x+y))]:

A := lhs(sys[1])+lhs(sys[2]);

st:=time(): Tuples(10, 4): time()-st;
                42.828

st:=time(): CartProdSeq([$0..4] $ 10): time()-st;
                46.593

@Markiyan Hirnyk

You are wrong on two points:

1. You forgot about the condition: acute triangle

2. This inequality requires symbolic, ie, the exact solution, rather than the approximate one.

@ThU 

You are right. Shoelace's formula is a wonderful formula! It is used in the text of the procedure  Area  for calculating the area of any region bounded non-selfintersecting line, if the whole or a part of the boundary is a broken line.

See  http://www.maplesoft.com/applications/view.aspx?SID=146470

@ThU 

You are right. Shoelace's formula is a wonderful formula! It is used in the text of the procedure  Area  for calculating the area of any region bounded non-selfintersecting line, if the whole or a part of the boundary is a broken line.

See  http://www.maplesoft.com/applications/view.aspx?SID=146470

@Carl Love

Work of your procedure:

P(3, 9);

           1

My procedure also works incorrectly if  N2 - N1<=1

@Carl Love

Work of your procedure:

P(3, 9);

           1

My procedure also works incorrectly if  N2 - N1<=1

To M. Hirnyk!

Of course, my answer is wrong. I partially solved by hand, and it seemed to me that  tan(tan^(-1)(4))=1/4

To C. Love!

To complete the solution should be checked, ie substitution of the answer to the original equation. The fact that the two equations  A=B  and  tan(A)=tan(B)  are not equivalent  (the second equation is a consequence of the first). Therefore, you may receive extraneous solutions.

1. 

U := [7, 9, 14]:

A := [-2, 7, 8, 12, 9, -78, 0]:

remove(x->is(x in U), A);

      [-2, 8, 12, -78, 0]

2.

C:=[[-2, 2], [7, 0], [8, 6], [12, 12], [9, 6], [-78, 45], [0, 7]]:

remove(x->is(x[1] in U), C);

         [[-2, 2], [8, 6], [12, 12], [-78, 45], [0, 7]]

3.

E := map(x->`if`(not x[1] in U, x, NULL), C);

         E := [[-2, 2], [8, 6], [12, 12], [-78, 45], [0, 7]]

minimize(3-4*cos(z)+cos(2*z), z = 0 .. infinity);

maximize(3-4*cos(z)+cos(2*z), z = 0 .. infinity);

minimize(3-4*t+2*t^2-1, t = -1 .. 1);  # Change  cos(z)=t

maximize(3-4*t+2*t^2-1, t = -1 .. 1);

PS.  It seems that Maple does not understand that there is a periodic function with period 2*Pi.

minimize(3-4*cos(z)+cos(2*z), z = 0 .. 2*Pi);

maximize(3-4*cos(z)+cos(2*z), z = 0 .. 2*Pi);

                                     0

                                     8