MaplePrimes - answers and comments on Question, trigonometric solve without floats

one way

pagan — Fri, 21 Oct 2011 04:45:04 Z

Granted, this involves some manually made choices for _B1 and _Z1 (the parameters of the solution, which you can run getassumptions against).

restart:

e := 4*cos(t)+2*cos(2*t)=0:

sol:=solve([e, 0 <= t, t <= 2*Pi], t,
            AllSolutions, Explicit):

#plot(lhs(e),t=0..2*Pi,tickmarks=[decimalticks,default]);

s1:=eval(convert({sol},`global`),[`_B1~`=0,`_Z1~`=0])
    union eval(convert({sol},`global`),[`_B1~`=0,`_Z1~`=1]):

s1:=select(type,map(rhs@op,s1),realcons):

select(z->is(z>=0 and z<=2*Pi), s1);

      /             /1  (1/2)   1\        /1  (1/2)   1\\ 
     { 2 Pi - arccos|- 3      - -|, arccos|- 3      - -| }
      \             \2          2/        \2          2// 

evalf(%);

                   {1.196061894, 5.087123414}

Another approach

Markiyan Hirnyk — Fri, 21 Oct 2011 11:06:57 Z

Because eval(...,[_B1~=0,_Z1~=0]) in the pagan's answer is made by hand, here is another approach:
> restart:
> e := 4*cos(t)+2*cos(2*t) = 0;
>sol := solve([e, 0 <= t, t <= 2*Pi], t, AllSolutions, Explicit);

{ t = arccos(1/2*sqrt(3) -1/2) + 2 Pi _Z1~ },

   { t = - arccos(1/2*sqrt(3) -1/2) + 2 Pi _Z1~ }, { t = Pi - arccos(1/2*sqrt(3) +1/2)


    - 2 _B1~ Pi + 2 _B1~ *arccos(1/2*sqrt(3) +1/2) + 2 Pi _Z1~ }
We see that sol[3] is complex:
>evalf(sol[3]);
{t = 3.141592654-.8314429458*I-6.283185308*_B1+(1.662885892*I)*_B1+6.283185308*_Z1}
Taking into account only real-valued solutions, we obtain
> with(RealDomain):
> iso2 := isolve(evalf(rhs(op(sol[2]))) <= evalf(2*Pi), evalf(rhs(op(sol[2]))) >= 0})

{_Z1~ = 1}
eval(sol[2], [op(iso2)]);
{t=- arccos(1/2*sqrt(3) -1/2) + 2 Pi}
> evalf(eval(sol[2], [op(iso2)]));

{t = 5.087123414}
> iso1 := isolve({evalf(rhs(op(sol[1]))) <= evalf(2*Pi), evalf(rhs(op(sol[1]))) >= 0});

{_Z1~ = 0}
>eval(sol[1], [op(iso1)]);

{t= arccos(1/2*sqrt(3) -1/2) }
> evalf(eval(sol[1], [op(iso1)]));

{t = 1.196061894}

extract_sols.mw

Thank you both

Andreas Madsen — Fri, 21 Oct 2011 13:54:08 Z

Thank you both. Since i use Math 2D I like Markiyans way. I also ended up createing this procedure to solve my problem:

> trigSolve := proc (eq, na) local sol, memory, restore, elems, i; sol := [solve([eq, 0

> trigSolve(4*cos(t)+2*cos(2*t) = 0, t) # {t = arccos((1/2)*sqrt(3)-1/2)}, {t = 2*Pi-arccos((1/2)*sqrt(3)-1/2)}

solve_proc.mw

I hope someone else will find it usefull, and perhaps post an improvement.

different view

Axel Vogt — Fri, 21 Oct 2011 18:45:52 Z

I would consider this as a polynomial equation in cos(t)
(and would work from there, perhaps not that different):

expand(eq);
[solve(%, cos(t))]; evalf(%);

                                        2
                     4 cos(t) + 4 cos(t)  - 2 = 0

Now one has to select the very one between -1 and +1
and arccos gives it.

Problem

Markiyan Hirnyk — Fri, 21 Oct 2011 08:44:19 Z

I cannot reproduce it. See solutions.mw

1D vs 2D

pagan — Fri, 21 Oct 2011 09:28:03 Z

@Markiyan Hirnyk The code posted for 2-argument eval to substitute B1~ and Z1~ is problematic in 2D Math input. The trailing tildes are being dismissed by the 2D Math parser.

I posted 1D Maple notation (plaintext) code. It is not always appropriate to paste such into the Standard GUI as 2D Math input. This is an example of that.

1D and 2D Maple are two different programming languages.