MATLAB: How to solve for x in an sin equation

symbolictrignometry

Hi MATLAB wizards

I am currently trying to solve the following equation; -7sin(x/7)=0 using symbolic algebra and the solve function in MATLAB.

The code that I used is as follows:

syms x

solve( (-7*sin(x/7)),x)

This code returns the value 0, which is one of the solutions, the other one is 7*pi.

So how can I produce both of these solutions?

Any form of help would be appreciated.

Best Answer

No. There are INFINITELY many solutions. So you cannot say you want the OTHER solution. You can simply enough find a positive solution though.

syms x positive
solve( (-7*sin(x/7)),x)
ans =
7*pi

Related Solutions

MATLAB: Using solve to solve a complex trigonometric equation

Hello Junha Kim,

syms x;
assume(x>=0 & x<=2*pi);
eqn = cos(x)*(1-sin(x)) == 0;
[solx, parameters, conditions] = solve(eqn, x, 'ReturnConditions',true);
assume(conditions);
restrictions = [x>=0, x<=2*pi];
solk = solve(restrictions, parameters);
valx = subs(solx, solk)

for more information, refer here

MATLAB: Solving trigonometric equation using “solve” function.

Whats the problem? There is no need to set any properties anyway on such a trivial problem.

syms x;
x = solve(230.769*cos(x)+153.846*sin(x) == 1.715, x);
vpa(x)
ans =
                                             2.1526153643152832650062014581097
 - 0.97661015722014818434287396730445 - 1.8367099231598242312011508394098e-40i

The second solution is real, it just has an infinitesimal imaginary part.

And of course, you can add any multiple of 2*pi to those solutions.

But even if you do as you did, then LOOK AT THE SOLUTION THAT YOU DID GET. THINK ABOUT IT!

xsol = solve(230.769*cos(x)+153.846*sin(x) == 1.715, x, 'IgnoreProperties',true, 'ReturnConditions', true)
xsol = 
  struct with fields:
             x: [2×1 sym]
    parameters: [1×1 sym]
    conditions: [2×1 sym]
>> vpa(xsol.x)
ans =
                                             6.283185307179586476925286766559*k + 2.1526153643152832650062014581097
 6.283185307179586476925286766559*k - (0.97661015722014818434287396730445 + 1.8367099231598242312011508394098e-40i)

Which says that for integer k, you can add any integer multiple of 2*pi to the solution. I'm pretty sure I said that initially.

xsol.conditions
ans =
 in(k, 'integer')
 in(k, 'integer')

Best Answer

Related Solutions

MATLAB: Using solve to solve a complex trigonometric equation

MATLAB: Solving trigonometric equation using “solve” function.

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