how to use solve

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alize beemiel
alize beemiel 2023-4-6
评论： alize beemiel 2023-4-6
hi ; i need help
I have this equation with 2 parametres a and b
syms b x a
solve ((2*cos(x) - b*(sin(x) + sin(a*x))) == 0) %chooses 'x' as the unknown and returns
I use solve but it return
Warning: Cannot find explicit solution.
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VBBV
VBBV 2023-4-6
移动：VBBV 2023-4-6
Ran in:
Consider these input values for a and b, for which solve function cant handle the solution. In such cases, use vpasolve to solve equation numerically as recommended by Matlab
syms x real
a = 4; % assume some value
b = 1.5; % assume value
sol=solve ((2*cos(x) - b*(sin(x) + sin(a*x))) == 0,[x]) %
Warning: Unable to solve symbolically. Returning a numeric solution using vpasolve.
sol = 
0.31726439340850840945560345483897
sol = vpasolve((2*cos(x) - b*(sin(x) + sin(a*x))) == 0,[x])
sol = 
0.31726439340850840945560345483897

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采纳的回答

VBBV
VBBV 2023-4-6
Ran in:
syms b x a
sol=solve ((2*cos(x) - b*(sin(x) + sin(a*x))) == 0,[x a b])
sol = struct with fields:
x: [2×1 sym] a: [2×1 sym] b: [2×1 sym]
%chooses 'x' as the unknown and returns
sol.x
ans = 
sol.a
ans = 
sol.b
ans = 
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alize beemiel
alize beemiel 2023-4-6
this is what i want by using matlab
alize beemiel
alize beemiel 2023-4-6
thank you Sir ..for your help and your time

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更多回答（1 个）

Chunru
Chunru 2023-4-6
Ran in:
There is no close form solution when a and b are arbitrary constant for the equation.
If you want to find the numerical solution of the equation with specified a and b, you can use vpasolve:
syms b x a
solve ((2*cos(x) - b*(sin(x) + sin(a*x))) == 0)
Warning: Unable to find explicit solution. For options, see help.
ans = Empty sym: 0-by-1
a = 2; b=0.1;
vpasolve ((2*cos(x) - b*(sin(x) + sin(a*x))) == 0) %chooses 'x' as the unknown and returns
ans = 
7.7984925986314595610243884059956
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Chunru
Chunru 2023-4-6
MATLAB tell you that its solver could not find the explicit solution.
alize beemiel
alize beemiel 2023-4-6
thank you for all
i will try to use Newton Raphson Methode maybe its give me an approximate solution

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