Solving for unisolated variable
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Hello all, I'm having a bit of trouble finding a method to solve a nonlinear equation. I've looked into "fsolve" a little but couldn't figure out if it would be effective. I'm using MatLab 2009.
Here is the equation, for ease I've simplified the constants to A,B,C. Variable to solve for is x.
x = arctan( (A - C*cos(x)) / (B*cos(x)) )
Which was derived from
A = B*sin(x) + C*cos(x)
Any help would be greatly appreciated... Rather than relying on MatLab to solve I would like to have x = F(A,B,C) where A,B,C are known but change with each iteration.
Another consideration is that this value needs to be found 5-10 times per second.
Thank you in advance! - Jon
回答（2 个）
Walter Roberson
2013-11-22
There are two solutions:
T0 = sqrt(C^2*B^2 - B^2*A^2 + B^4);
X = [(C*A + T0)/(C^2+B^2), (C*A - T0)/(C^2+B^2)]; %just sign difference between two
Y = (A - C*X) / B;
atan2(Y, X)
Roger Stafford
2013-11-23
Two solutions can be expressed as:
x = asin(A/sqrt(B^2+C^2)) - atan2(C,B);
and
x = pi - asin(A/sqrt(B^2+C^2)) - atan2(C,B)
Also either value with any integral multiple of 2*pi added or subtracted is a solution, thus giving infinitely many solutions.
3 个评论
Jonathan
2013-11-23
Roger Stafford
2013-11-23
My reasoning went this way. Starting with
A = B*sin(x) + C*cos(x)
notice that any pair of numbers B and C can always be represented by
B = sqrt(B^2+C^2)*cos(t)
C = sqrt(B^2+C^2)*sin(t)
for an appropriate angle t, which can be evaluated by
t = atan2(C,B)
(That in fact is one way to define the function atan2.) This gives
A = sqrt(B^2+C^2)*(cos(t)*sin(x)+sin(t)*cos(x))
= sqrt(B^2+C^2)*sin(x+t)
sin(x+t) = A/sqrt(B^2+C^2)
x + t = asin(A/sqrt(B^2+C^2))
x = asin(A/sqrt(B^2+C^2)) - t = asin(A/sqrt(B^2+C^2)) - atan2(C,B)
Jonathan
2013-11-23
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