@tomleslie  The same can write shorter:

f:=x->x*cos(x)-sin(x)*sin(x/1000);
r:= 0:
to 20 do
r:=RootFinding[NextZero](f, r);
od;

@snowww  

restart;

N := {F = 1, K = 1.2}:

Eq := diff(C(x),x,x)-F*(diff(C(x),x))-F*K*C(x) = 0:

Bcs := F*D(C)(0) = C(0)-1, D(C)(1) = 0:

Sol := dsolve(eval({Bcs, Eq}, N), C(x)):

assign(%);

plot([C(x), eval(exp(-K*x), N[2])], x = 0 .. 3, 0 .. 3, color = [red, blue], thickness = 2);

@snowww  You have to love Maple and work with it constantly.

@farzane  The above code works well in this case also:

A:= .5464691235-.4473247264*I,  -.4563184747+1.*10^(-14)*I,  .5464691235+.4473247264*I, 0.9091925189:

Re(sort([A], (z1,z2)->abs(Im(z1))<abs(Im(z2)))[1]);

                                    0.9091925189

@Markiyan Hirnyk  You're right of course. Think of it as a joke!

@Markiyan Hirnyk 

P:=a*b^2+sqrt(41)*b*c-a^3*c^2;

P1:=subs(sqrt(41)=d, a*b^2+sqrt(41)*b*c-a^3*c^2);

Q:=expand(P1^7);

ind:=indets(%) minus {d};

n:=igcd(seq(degree(Q,s), s=ind));

subs(d=sqrt(41), RealDomain[simplify](Q^(1/n), symbolic));

@Markiyan Hirnyk 

1) I'm not familiar enough with the theory of polynomial ideals and radicals in them. As I understand it from the original post, it is necessary for a given polynomial in several variables over real numbers just to find the root from it of maximum order, of course, if this root there exists.

2)  I did not compare my method with other methods of solving this problem.

3) I just answered that question OP   "Is there another commands for this?"

@MDD  This means that an assumption is imposed on the variable a. If you want the symbol tilda to be absent, click

Tools->Options->Display->Assumed variables->No Annotation-> Apply Globally

@MDD  You must (in the copiable form) provide the full text of the code in which this error occurs.

@MDD   it is.

@Markiyan Hirnyk 

A := [x, y, x^2*y, x*y^2, y^2]:

B := [x^2, y^3]:

remove(c->`or`(seq(divide(c, b), b = B)) , A);

                                [x, y, x*y^2, y^2]

@Carl Love  Thank you very much for your response. This is a very effective way!

@Markiyan Hirnyk   for your interest! I meant that a segment is a closed interval. Of course easily to adjust procedure for all cases.

@Carl Love  I do not understand why on your images  additional lines appears, for example for k=1 .  Compare with my image:

F:=(x,y)->(1/2*x^2-3/2*y^3,-1/2*x^3+1/3*y^2):  # Nonlinear mapping R^2->R^2

X1:=t->(t,0): X2:=t->(0,t): X3:=t->(t,1): X4:=t->(1,t):

plot([[X1(t),t=0..1], [X2(t),t=0..1], [X3(t),t=0..1], [X4(t),t=0..1], [F(X1(t)), t=0..1], [F(X2(t)), t=0..1], [F(X3(t)), t=0..1], [F(X4(t)), t=0..1]], color=[red$4,blue$4], thickness=3, scaling=constrained);

@MDD  Your procedure was not working properly. What does  lt(p, T)  mean in it? An infinite loop occurs.