Numerical solution of a trigonometric equation

I want to solve the trigonometric equation, so that for each $x$ and $k$ we will have either 0,1 or 2 solutions.

sin(k+a)=x*cos(k)

$x$ is a variable, $a$ is real, (a is argument of a complex number) so I take $0<=a<=pi$ and similarly for $k$ I have the condition $0<=k<=pi$. How to solve it numerically? I am unsure whether to use 'solve' or 'vpasolve' or another method. Secondly, I want to visualize the solutions by creating a table of solutions [x,k] and by plotting them. I would do the following in Mathematica

eq[a_, k_, x_] := Sin[k + a] == x*Cos[k];
sol[x_?NumericQ, k_ /; 0 <= k <= Pi] := a /. 
NSolve[{eq[a, k, x], 0 <= a <= Pi}, a, Reals]

and then e.g., sol[0.1,0.1](=2.94193) would give me solution at x=0.1, k=0.1, and so on. Meaning that e.g., for each 'x', we can plot 'a' vs 'k'. and for each 'x' and 'k', we have a number of solutions either 0,1 or 2 etc. How to do it in MATLAB?

[Torsten]

Why do you solve for x if you want to solve for a ?

[AtoZ]

You are right. My mistake in copy pasting from help. How do i do it here? I have to predefine x, a, k? Means a range for all these values? actually x is different in that it has a range from 0 to 5.

[Torsten]

syms x k a
eqn = sin(k+a)==x*cos(k);
sol = solve(eqn,a)

does not work ?

[AtoZ]

it gives an answer

asin(x*cos(k)) - k
 pi - k - asin(x*cos(k))

I wanted to see the numerical results, preferably as a table/and or plot. So where can I give the range of values to the parameters?

[AtoZ]

e.g., for each x, we can plot a vs k. and for each x and k, we have solutions either 0,1 or 2.

Answer 1

[Torsten]

function main
  a = fun(0.1,0.1)
end

function a = fun(x,k)
  t = x*cos(k);
  if abs(t) <= 1
    sol(1) = asin(t) - k;
    sol(2) = pi - k - asin(t);
    a = sol(sol >=0 & sol <= pi);
  else
    a = [];
  end
end

[AtoZ]

@Torsten Thanks. It is giving an error

Function definitions are not permitted in this context.

[Torsten]

Put the above code in a file called "main.m", load the file and run it.

[AtoZ]

@Torsten thanks. It seems to be a rather complicated code to an apparently simple trigonometric equation. Isn't there a simpler solution to the problem? Using vpasolve or any other workaround? As I wrote earlier, I could do it with a simple MMA code and Matlab being more powerful than MMA, I presume there should be simpler ways of doing it.?

[Stephen23]

"It seems to be a rather complicated code to an apparently simple trigonometric equation."

Are you wanting to solve an equation, or get a numeric solution?

These are two totally different tasks that use totally different methods. Finding roots and similar tasks can be done very efficiently, probably much more efficiently than messing around with symbolic maths trying to solve equations. But only you can decide what you need to do.

"Isn't there a simpler solution to the problem? Using vpasolve or any other workaround?"

You seem to be confusing conceptually simple with not writing much code. These two statments are not necessarily synonymous, and to think that they are is a misleading approach to coding. Torsten's approach is a conceptually simple numeric solution using quite basic commands. The fact that it takes a few lines does not change this. In contrast your approach using vpssolve, or your question and answer here

https://www.mathworks.com/matlabcentral/answers/428266-two-for-loops-taking-very-long-to-get-the-complete-solutions

show that you seem determined to solve this using slow symbolic methods, simply because you can write less. Sure, if you believe that writing one command is "simple", then go for it. But your request for "simplicity" ignores the complexity that is hidden behind that command, which is certainly not simple and for a numeric output is (depending on the problem) nowhere near as efficient as well written numeric code. Do not be surprised that one "simple" vpasolve command (which is not really simple at all, as it works with symbolic maths) is slower than several truly simple low-level numeric operations (supported by your computers hardware and possibly some highly optimized C/Fortram modules....).

Numeric and symbolic calculations are really totally different animals. They are not comparable by judging them on "how much I have to write". If you don't want to write much, then use slow vpasolve. If you want fast code, then learn from highly experienced Torsten who can show you how to write efficient numeric code.

[AtoZ]

@Stephen, Thanks. I understand your point. I was just a bit worried why did we need all that code, and how to generalize those conditions (code) for situations not as simple as this equation. I think it is well explained to me now. I was not on a right track.

[AtoZ]

@Torsten could you please tell how to get the output as a two dimensional array in the form of a table? So its easier to identify regions with multiple solutions. I tried

T=table(x{:},k{:})

but it doesn't seem to work.

[Torsten]

k = 0:0.1:1;
x = 0:0.1:1;
T = table(k,x);
T.k(6)   %=k(6)
T.x(4)   %=x(4)

[AtoZ]

@Torsten Much appreciated, Thanks.

[AtoZ]

@Torsten, sorry to bother again. The prescription of dot indexing that you wrote, only gives the variable indexing, i.e., it returns only the values

0 0.1 0.2 0.3 ... 0.9 1.

two times. How is it tied to the solutions? Here

clear all
syms k a x
f = sin(k+a) - x*cos(x)==0;
solution=solve(f,a,[0 pi])
x=0:0.1:1;
k=0:0.1:1;
solutions=vpa(subs(solution),3)
T = table(k,x);
T.k 
T.x

[Torsten]

function main
  x = 0:0.1:1;
  k = 0:0.1:1;
  [X,K] = meshgrid(x,k);
  X = X(:);
  K = K(:);
  for i=1:numel(K)
    a(i,:) = fun(X(i),K(i));
  end
  T = table(X,K,a);
end

function a = fun(x,k)
  t = x*cos(k);
  if abs(t) <= 1
    sol(1) = asin(t) - k;
    sol(2) = pi - k - asin(t);
    a = sol(sol >=0 & sol <= pi);
  end
  a(end+1:2) = NaN;
end

[AtoZ]

@Torsten +1 for this.. Thanks..

[Bruno Luong]

Careful, this line is buggy in Torsten's code

      t = x*cos(k);

[Torsten]

Thanks, Bruno, but equation was here

sin(k+a) = x*cos(k)

I think the OP just wants to experiment a bit with MATLAB.

[Bruno Luong]

Ah OK.