Numerical solution of a trigonometric equation

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AtoZ 2018-11-5

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/428022-numerical-solution-of-a-trigonometric-equation

编辑： Torsten 2018-11-8

采纳的回答： Torsten

在 MATLAB Online 中打开

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?

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AtoZ 2018-11-5

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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 2018-11-5

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.

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采纳的回答

Torsten 2018-11-5

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/428022-numerical-solution-of-a-trigonometric-equation#answer_345255

编辑：Torsten 2018-11-5

在 MATLAB Online 中打开

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

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Stephen23 2018-11-6

编辑：Stephen23 2018-11-7

"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.