Here is the solution of the logic problem for composition of departments:

restart;

Names:={Alex, Betty, Carol, Dan, Earl, Fay, George, Harry}:

A:=combinat[choose](Names, 3):

L:=[]:

for i in A do

B:=combinat[choose](Names minus i, 3);

for j in B do

L:=[op(L), [i, j, (Names minus i) minus j]];

od: od:

Organization:={seq(op(combinat[permute](L[k])), k=1..nops(L))}:

Terms:=[Dan in Administration, Fay in Personnel, Alex in Personnel, 'nops(Personnel)=2', not(Earl in Administration), not(Harry in Administration), not(Carol in Marketing), not(George in Administration)]:

Departments:=[]:

for variant in Organization do

Personnel:=variant[1]: Administration:=variant[2]: Marketing:=variant[3]:

if convert(Terms, `and`) then Departments:=[op(Departments), variant]; fi;

od:

op(Departments);

                           [{Fay, Alex}, {Dan, Betty, Carol}, {Earl, George, Harry}]

Thus we have the unique solution:

Personnel={Fay, Alex},  Administration={Dan, Betty, Carol},  Marketing={Earl, George, Harry}

Similarly, we can find the distribution of sorts of sport.

In fact, there is no error, because  cos  function is even, ie  cos(-x) = cos(x). If you do not confuse the extra parentheses, you can write:

cos(``(Phi1(x)-psi));

If further transformations are not planned, you can just write:

cos(``*Phi1(x)-psi);

Replace simplify by  radnormal .

Procedure  Horner  using Horner's method calculates the value of the polynomial  P  in the point  x0 :

restart;

Horner:=proc(P::polynom, x0::realcons)

local n, x, b, i, a;

n:=degree(P); x:=op(indets(P));

assign(seq(a[k]=coeff(P, x, k), k=0..n)); b:=a[n];

for i from n-1 by -1 to 0 do

b:=a[i]+b*x0;

od;

b;

end;

Example:

Horner(x^3-5*x^2-3*x+4, 3);

                   -23

Example:

f:=unapply(expand(sum((x1+x2+x3)^i, i=1..5)), x1,x2,x3);

f(a,b,c);

restart;

sol := solve([sin(a)-sin(b) = 0, b >= 0, b < 2*Pi], b, allsolutions, explicit);

seq(eval(op(sol[i]), op(`minus`(indets(sol[i]), {a, b})) = 0), i = 1 .. 2);

 Addition:  if instead of  a  specific angle to write, there is no problem.

restart;

solve({sin(Pi/6)-sin(b) = 0, b>=0, b<2*Pi}, b, allsolutions, explicit);

You can just use  simplify  command. For a more substantial simplification seems to be  specified numerical values ​​to all constants.

simplify(dsolve({ODE,bcs}));

SeqInColumn:=proc(S)

local i;

for i to nargs-1 do

print(args[i], ``);

od;

args[nargs];

end:

Your example:

B := b=2:

C := c=3:

A := B, C:

SeqInColumn(A);

                b=2,
                 c=3

Your example:

restart;

eq:=f(x)=1+x*f(x)^2;

f(x):=a+b*x+c*x^2+d*x^3;

expand(eq);

sys:={seq(coeff(lhs(eq), x, k)=coeff(rhs(eq), x, k), k=0..3)};

solve(sys);

Or

member(5, {seq(13^k mod 100, k=1..1000)});

                                   false

The benefit a set compared to a list that automatically repetitive elements removed from a set and it is sorted in ascending order.

Try

 Hits := identify(d);

for i from 1 to 501 do if type(Hits[i], 'float') = false then  print(Hits[i]) end if end do;

The law of sines:

AB/sin(C)=BC/sin(A)=AC/sin(B)=2*R=6

We have  sin(C)=AB/6=sqrt(2)/2,  sin(A)=BC/6=sqrt(2)/3

Two cases are possible:  C=Pi/4  or  C=3*Pi/4

Uniquely  A=arcsin(sqrt(2)/3)  because  A<C

Obviously that  cos(B)=cos(Pi - (A+C))=-cos(A+C)  

Next we use Maple and the law of cosines:

c1:=expand(-cos(Pi/4+arcsin(sqrt(2)/3)));

c2:=expand(-cos(3*Pi/4+arcsin(sqrt(2)/3)));

AB:=3*sqrt(2): BC:=2*sqrt(2):

radnormal(sqrt(AB^2+BC^2-2*AB*BC*c1));  # first answer

radnormal(sqrt(AB^2+BC^2-2*AB*BC*c2));  # second answer

Addition - visualization of both solutions:

expand(2*sin(x+(1/4)*Pi));
``*coeff(%, sin(x))*sin(x)+``*coeff(%, cos(x))*cos(x);

Addition:  In Standard M 16 the same result can be obtained  simpler

 expand(``*2*sin(x+(1/4)*Pi));

subs(2=`2`, expand(2*(sin(x+Pi/4))));

Likely this can only be done numerically:

f := BesselJ(1, t)*(z*BesselJ(0, t)*BesselJ(1, z)-t*BesselJ(0, z)*BesselJ(1, t))/((1+10*t)*(z^2-t^2)):

F:=unapply(Int(f, t=0..infinity), z);

evalf([F(0), F(1), F(2)]);