MATLAB help (finding all the possible values for x)?

Question

Tristan 2013-9-5

1
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/86431-matlab-help-finding-all-the-possible-values-for-x

采纳的回答： Shashank Prasanna

在 MATLAB Online 中打开

Hi, I'm new to MATLAB,

I need to find all the possible values for x knowing that:

0<x<10

and

1.8*cos(1.8*x)+1.2*cos(1.2*x)=0

the only thing that works is

>> evalin(symengine, 'numeric::solve(1.8*cos(1.8*x)+1.2*cos(1.2*x)=0, x = 0..10)')
ans =
1.0057538744094438603875879163721

but it only gives one of the possible answers.

I should be getting

x≈

1.00575

2.97903

4.6645

5.80748

7.49295

9.46622

thanks

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Answer 1

Shashank Prasanna 2013-9-5

1
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https://ww2.mathworks.cn/matlabcentral/answers/86431-matlab-help-finding-all-the-possible-values-for-x#answer_95900

在 MATLAB Online 中打开

syms x
x = solve(1.8*cos(1.8*x)+1.2*cos(1.2*x))

This gave me the analytical solution.

A more feasible way would be to solve this numerically using MATLAB:

x = fsolve(@(x)1.8*cos(1.8*x)+1.2*cos(1.2*x),10*rand(1))

Each time you run this you will get one of the different results. The second argument asks FSOLVE to start solving from a different start point.

http://www.mathworks.com/help/optim/ug/fsolve.html

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Answer 2

Roger Stafford 2013-9-5

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/86431-matlab-help-finding-all-the-possible-values-for-x#answer_95911

Another approach is to expand both cosines in terms of cos(.6*x) which gives a cubic equation in the quantity. Any of its roots which are real and between -1 and +1 will give you an arccosine equation from which you can find all the solutions in the range you state.