# solving implicit equation

Hi experts!

I want to resolve the equation: $A.sin(x)=B.(C-x)$

where A, C and B are constants, $0\leq x\leq \pi$, $A$ and $B$ are fixed, and $0\leq C\leq \pi$.

I want to obtain the analytic solution, $x=f(C)$, (if exist) for different values of $C$ and if only implicit solution exist, obtain the curve and and table with the values of:

$x$ vs. $C$

How can I do that?

Thanks a lot!

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One option, and this might be suboptimal, is to replace sin(x) with its taylor series.

Because

sin(x) = x - (1/6)x^3 + (1/120)x^5 - (1/5040)x^7 + (1/362880)x^9 - O(x^11)

we would change

Asin(x) = BC - B*x

Ax - (A/6)x^3 + (A/120)x^5 - (A/5040)x^7 + (A/362880)x^9 - O(A x^11) = BC - Bx

giving you then

-BC + (A+B)x - (A/6)x^3 + (A/120)x^5 - (A/5040)x^7 + (A/362880)x^9 - O(A x^11) = 0

or if you want a few more terms

-BC + (A+B)x - (A/3!)x^3 + (A/5!)x^5 - (A/7!)x^7 + (A/9!)x^9 - (A/11!)x^11 + (A/13!)x^13 + O(A x^15) = 0

This does reveal that varying C will only affect the answer by a constant.

If A and B were fixed constants, known in advance, then you could use "for" and "find_root" to find a value of x for a long sequence of C values, forming a table of values.

For example:

http://sagecell.sagemath.org/?z=eJxtk...

more

Thanks. It can't be done wothout using Taylor expansion?

( 2014-07-20 17:36:54 -0600 )edit

I'm guessing that you need either the Taylor expansion or some competitor of it, like the Chebychev polynomials. The main issue is that you have both sin x and a linear polynomial in x. Therefore, I really can't see how you would "solve for x." I'm about 98% sure but I don't want to say 100%, because mathematics is always full of surprises.

( 2014-07-21 12:15:59 -0600 )edit