Multiple Answers From 'solve'

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Zachary
Zachary 2012-1-9
I have an equation that I am passing to the 'solve' function. I know for a fact that it has two solutions. It only returns one solution. This is how I'm calling solve:
solve(S)
where S is a string containing the equation. It's pretty ugly but I can provide it if needed. What I need is a way to get that second solution.
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Andrew Newell
Andrew Newell 2012-1-9
Are you trying to solve symbolically for theta4?
Zachary
Zachary 2012-1-9
Yes, I'm feeding the solve line a series of theta2 values in a loop. I am returned a symbolic answer for theta4 which I turn into a double and plot.

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Andrew Newell
Andrew Newell 2012-1-9
syms theta2 theta4
S=(-4.19*sin(theta2*pi/180)-5.166*sin(theta4*pi/180)+23.3*sin(82.99*pi/180))^2+(-4.19*cos(theta2*pi/180)-5.166*cos(theta4*pi/180)+23.3*cos(82.99*pi/180))^2-24.073^2;
sols = solve(S,theta4);
sols = subs(sols,theta2,73.5);
sols =
[-115.9829194]
[ ]
[-73.90876566]
The first solution, added to 180, gives 64.0170806.
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Zachary
Zachary 2012-1-9
Upon playing with it a bit and trying other test cases, I've found something interesting.
One of the two answers is right, plus or minus 180. Also, after this little bit of post-processing, it happens that the answer I want is always the only one with zero imaginary component.
So I just need to write some logic to figure out which one I want.
You've been extremely helpful, many thanks for the assistance!
Walter Roberson
Walter Roberson 2012-1-9
The reason for the +/- 180 is shown in my answer.

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更多回答（1 个）

Walter Roberson
Walter Roberson 2012-1-9
Take your original equation,
24.073^2 = (-4.19*sin((1/180)*theta2*Pi) - 5.166*sin((1/180)*theta4*Pi) + 23.3*sin(.4610555556*Pi))^2 + (-4.19*cos((1/180)*theta2*Pi) - 5.166*cos((1/180)*theta4*Pi) + 23.3*cos(.4610555556*Pi))^2
and do trig substitutions on it.
24.073^2 = 587.1336560 + 43.29108000*cos((1/180)*theta2*Pi - (1/180)*theta4*Pi) - 195.2540000*cos((1/180)*theta2*Pi-.4610555556*Pi) - 240.7356000*cos((1/180)*theta4*Pi-.4610555556*Pi)
Substitute in any one value for theta2 and the equation becomes one involving constants, and cos() of theta4 converted from degrees to radians. It then becomes obvious that the solution must be periodic over 360 (degrees) and the positive-going and negative-going solutions must be 180 apart.
Thus, all you need from solve() is a single solution, which you can then normalize according to your preferred range, with the other solution in that preferred range 180 apart.
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Walter Roberson
Walter Roberson 2012-1-9
Use vpa() or double() on the expression
Note: above I used Pi with capital-P . In MATLAB you would use pi with lower-case-p .
Zachary
Zachary 2012-1-10
I get the same single solution I got before... that's some clever trig though. Andrew seems to have figured out how to make it give me both solutions.

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