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

@Markiyan Hirnyk 

1. Of course, you are right in that. I just do not understand why any problem you want to solve exclusively with Maple. It's interesting, but not always effective. Even if the result is obtained by using Maple, it should be tested as sometimes bugs are possible.

2. For  x = -1  it is obvious that the total sum is 0 since each term of the series is 0. It can also be set with Maple, but it requires some effort. The fact is that Maple does not understand that  3^n  is odd for any non-negative integer. For the success the first term of the series should be calculated separately. See

is((3^n)::odd) assuming n::nonnegint;

                           FAIL

eval((x^(3^n)+(x^(3^n))^2)/(1-x^(3^(n+1))), {n = 0, x = -1})+eval(sum((x^(3^n)+(x^(3^n))^2)/(1-x^(3^(n+1))), n = 1 .. infinity), x = -1) ;

                              0

@Markiyan Hirnyk 

1. Of course, you are right in that. I just do not understand why any problem you want to solve exclusively with Maple. It's interesting, but not always effective. Even if the result is obtained by using Maple, it should be tested as sometimes bugs are possible.

2. For  x = -1  it is obvious that the total sum is 0 since each term of the series is 0. It can also be set with Maple, but it requires some effort. The fact is that Maple does not understand that  3^n  is odd for any non-negative integer. For the success the first term of the series should be calculated separately. See

is((3^n)::odd) assuming n::nonnegint;

                           FAIL

eval((x^(3^n)+(x^(3^n))^2)/(1-x^(3^(n+1))), {n = 0, x = -1})+eval(sum((x^(3^n)+(x^(3^n))^2)/(1-x^(3^(n+1))), n = 1 .. infinity), x = -1) ;

                              0

You can see that Maple is unable to find the sum, even taking into account the identity  a[n]=1/(1-x^(3^n)) - 1/(1-x^(3^(n+1))). It needs a little help.

From the identity follows that the partial sum of the series is  S_n = 1/(1-x) - 1/(1-x^(3^(n+1))) . However, even such a simple limit Maple does not finds. Only if we make a change  N=3^(n+1) , then the problem will be solved.

S_n := 1/(1-x)-1/(1-x^(3^(n+1))):

S_N := subs(3^(n+1) = N, S_n):

limit(S_N, N = infinity)  assuming  x::real, abs(x) < 1;

limit(S_N, N = infinity)  assuming  x::real, x < -1;

limit(S_N, N = infinity)  assuming  x::real, x > 1;

You can see that Maple is unable to find the sum, even taking into account the identity  a[n]=1/(1-x^(3^n)) - 1/(1-x^(3^(n+1))). It needs a little help.

From the identity follows that the partial sum of the series is  S_n = 1/(1-x) - 1/(1-x^(3^(n+1))) . However, even such a simple limit Maple does not finds. Only if we make a change  N=3^(n+1) , then the problem will be solved.

S_n := 1/(1-x)-1/(1-x^(3^(n+1))):

S_N := subs(3^(n+1) = N, S_n):

limit(S_N, N = infinity)  assuming  x::real, abs(x) < 1;

limit(S_N, N = infinity)  assuming  x::real, x < -1;

limit(S_N, N = infinity)  assuming  x::real, x > 1;

You need to give a concrete and precise formulation  of your problem. Only in this case, you can expect to receive useful advice.

I beg your pardon! I did not notice the condition "has four distinct solutions "

eq:= x^4 -(3*m+2)*x^2 + 3*m+1:

sol:=[solve(eq=0,x)]:

L:=[seq(1..nops(sol))]:

op(map(allvalues,[solve([seq(sol[i]<2, i=1..nops(sol)),seq(seq(`if`(i<>j, sol[i]<>sol[j], NULL),j in L), i in L)],m)]));

                                                {-1/3 < m, m < 0}, {0 < m, m < 1}

I beg your pardon! I did not notice the condition "has four distinct solutions "

eq:= x^4 -(3*m+2)*x^2 + 3*m+1:

sol:=[solve(eq=0,x)]:

L:=[seq(1..nops(sol))]:

op(map(allvalues,[solve([seq(sol[i]<2, i=1..nops(sol)),seq(seq(`if`(i<>j, sol[i]<>sol[j], NULL),j in L), i in L)],m)]));

                                                {-1/3 < m, m < 0}, {0 < m, m < 1}

@Carl Love

If you want to see the gridlines on the filled region then  transparency  option can be used. 

@Carl Love

If you want to see the gridlines on the filled region then  transparency  option can be used. 

According to Markiyan's idea:

l1:=[a1*t+x1, b1*t+y1, c1*t+z1]:

l2:=[a2*s+x2, b2*s+y2, c2*s+z2]:

minimize(add((l1[i]-l2[i])^2, i=1..3), s=-infinity..infinity, t=-infinity..infinity) assuming a1^2+b1^2+c1^2>0 and a2^2+b2^2+c2^2>0:

RealDomain[simplify](sqrt(%)); 

@Markiyan Hirnyk 

Let the questioner will decide himself which option is more suitable for him. Purely terminological distinction.

@Markiyan Hirnyk 

Let the questioner will decide himself which option is more suitable for him. Purely terminological distinction.