Where is the problem?

a := 2: b := 10: c := -4: d := 5: rs := 4:
A:={subs(filler=ln, filler(a*x+b)-filler(c*x+d) = rs)};
B:=map2(proc(x) x>0 end@op,1, indets(A, 'ln(dependent(x))'));
solve(A union B);
subs(%, A);
evalf(indets(%,function));

You can reason the use of normal, but factor always looks better.

A:=((x^3+a*x+b) -(y^3+a*y+b) )^2/(x-y)^4;
factor(A), normal(A);
[(factor@numer, factor@denom)(A)], [(normal@numer, normal@denom)(A)];
G:=gcd((numer,denom)(A));
[factor(numer(A)/G), factor(denom(A)/G)], [normal(numer(A)/G), normal(denom(A)/G)];

F := proc(E, V::{list, set})
     local T := 'exp(anything)';
       factor(subsindets(E, T, expand));
       `@`(proc(E, R) combine(E, power)*collect(combine(R, power), V, factor) end proc, selectremove)(type, %,  {T, T^anything});
       subsindets(%, T, factor);
     end proc;

S := {Y = (nu__p+1)*(nu__p-3), Z = (nu__p+1)^2};

subs(S , F(simplify(Ty4, S, [nu__p, Y, Z]), [Y, Z, exp, kappa, omega]));

gives:

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Try:
map(((x -> x) = expand)@cos ,[5*t, 6*t]);
 

Edit:
You can use ?solve,identity like Carl Love pointed out, or you can try this brute force method:
A := cos(t)^6+a*cos(t)^4*sin(t)^2+b*cos(t)^2*sin(t)^4+c*sin(t)^6 = cos(6*t);
combine((lhs-rhs)(A));
solve({coeffs}(%, indets(%, dependent(t))));

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That should be
fprintf(output, "%f\t %f\t \n", x, evalf(b)) ;

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There is an entire worksheet in that one line. It is recoverable I believe.

I've selected the entire contents of the sheet and converted it to Plain Text.
Result attached. corrupted-5.mw

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This curve is well known, parameterizable curve. You can do the following:

VR := -3 .. 3, -3 .. 3;

-3*a*x*y + x^3 + y^3 = 0;
Q1 := algcurves:-parametrization((lhs - rhs)(%), x, y, t);
plot(eval([op(Q1), t = 0 .. infinity], a = 1));
plot(eval([op(Q1), t = -1 .. 0], a = 1));
plot(eval([op(Q1), t = -infinity .. -1], a = 1));
plot(eval([seq([op(Q1), t = R], R = [-infinity .. -1, -1 .. 0, 0 .. infinity])], a = 1), color = red, view = [VR]);

x + y + a = 0;
Q2 := algcurves:-parametrization((lhs - rhs)(%), x, y, t);
plot(eval([op(Q2), t = -infinity .. infinity], a = 1), color = blue, view = [VR]);

plots[display](
plot(eval([seq([op(Q1), t = R], R = [-infinity .. -1, -1 .. 0, 0 .. infinity])], a = 1), color = red, view = [VR]), 
plot(eval([op(Q2), t = -infinity .. infinity], a = 1), color = blue, linestyle=dash, view = [VR])
);

See if this helps:
Go to:
Tools > Options > Display
and checkmark : Always insert new execution group after executing

Inside of f1 you could use:
if is(x, EvalfableProp) = true then "Your Test Goes Here" else 'procname(args)' fi
also define
D(f1):=0;

If you can anticipate the answer then this may be fastest method:
mtaylor(f1, [x, y], 24);
#or poisson(f1, [y], 24);
gfun[seriestolist](series(%, y, 24)):
gfun[guessgf](%, y, [ogf]);

a:=map(`^`, [8, 9 ,9 ,7 ,9 ,10 ,5], -1) mod 11;

Using Your suggestion I managed the code below. Nothing else worked.

G := (x, X) -> map((f, x, X) -> f(x) = subs(exp(I*x) = X, expand(convert(f(x),exp))), [sin, cos], x, X);
H := (x, X) -> X = convert(exp(I*x), trig);
expand(evala(subs(G(x, X), G(y, Y), ((expand@numer)/(expand@denom))(f1))));
simplify(subs(H(x, X), H(y, Y), %), trig);

Your desired solution differs from the general one as illustrated:
map[3](applyop, S->remove(evalb, S), 1, [eliminate(eqns union {theta[2, 1] = 1, theta[3, 0] = beta}, fc)]);

It's only a subcase (or so I believe).

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alias(mt=LinearAlgebra[Transpose], mht=LinearAlgebra[HermitianTranspose]);
map(radnormal, evalm(mht(F) &* F));
map(radnormal, evalm(mt(F) &* F));

Is this the issue:

alias(mt = LinearAlgebra[Transpose], mht = LinearAlgebra[HermitianTranspose]);
map(radnormal, evalm(mht(F) &* F));
map(radnormal, evalm(mt(F) &* F));

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