Why isn't this solve working?

Question

Joshua D'Agostino 2015-1-20

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/170627-why-isn-t-this-solve-working

评论： Star Strider 2015-1-20

采纳的回答： Star Strider

在 MATLAB Online 中打开

I'm trying to get out 2 values for an angle, gamma:

gamma = asin((sqrt(1 + d^2)*sin(a))/r)

is the formula for the angle. I understand that asin simply returns the principle value, and so if I want to get out 2 values I's have to solve sin(gamma) = ... etc.

I just can't get it to work. Here's my code:

syms gamma
r = 0.3584;
d = 0.6342;
l = degtorad(10);
b = degtorad(40);
a = (b - atan(d))^2 + l^2 % just how it's defined; these are all just details anyway
S = solve('sin(gamma) = (sqrt(1 + d^2)*sin(a))/r',gamma)

and here is the output:

Error using solve>processString (line 354)
' sin(gamma) = (sqrt(1 + d^2)*sin(a))/r ' is not a valid expression or equation.
Error in solve>getEqns (line 284)
      eqns = processString(eqns, v, vc);
Error in solve (line 160)
[eqns,vars,options] = getEqns(varargin{:});

How is that not a valid expression? I don't know where the problem is.

Thanks in advance for all help!

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Answer 1

Star Strider 2015-1-20

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链接

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https://ww2.mathworks.cn/matlabcentral/answers/170627-why-isn-t-this-solve-working#answer_165323

在 MATLAB Online 中打开

This works (in R2014b) with no other changes in your code (other than my having to define my own deg2rad function):

degtorad = @(x) x.*pi/180;
S = solve(sin(gamma) == (sqrt(1 + d^2)*sin(a))/r,gamma)

producing:

S =
        asin(2863908454326725/18014398509481984)
   pi - asin(2863908454326725/18014398509481984)

or with:

S = vpa(S)

produces:

S =
   0.15965626771650156828855864052742
    2.9819363858732916701740847427521

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Joshua D'Agostino 2015-1-20

right, I'm just an idiot. Thanks so much!

Star Strider 2015-1-20

My pleasure!

You’re definitely not an idiot! None of us learned MATLAB by osmosis. We all have our share of error messages, crashed code, and logic statements that did exactly the opposite of what we intended. (I do, anyway!) Keep coding, and share your expertise here when you have the opportunity.

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