Solve 2*C*sin(W)+P*cos(N*W)=P for W

[SSGD]

Is there a way to solve for W in the below equation. There has to be multiple solution for W, but I am at a loss as to how to solve this.

2*C*sin(W)+P*cos(N*W)=P or 2*C/P*sin(W)+cos(N*W)=1

C and P are constants
N is an integer

 

[mfb]

Not in a closed form for general N. There are special cases where a solution is possible, e.g. N=1.

I would expect N=2 and maybe N=3 and N=4 to have exact solutions as well, but not larger N. Careful: I didn't check those cases in detail.

You can write sine and cosine as sum of two complex exponentials, and convert your expression to a polynomial (with powers of e^iW). Those polynomial equations have proper, general exact solutions only up to 4th order.

 

[SSGD]

Made an error... change the sign before the P from + to -

2*C*sin(W)-P*cos(N*W)=P or 2*C/P*sin(W)-cos(N*W)=1

 

[mfb]

That doesn't matter. C and P can be arbitrary constants anyway.

 

[SSGD]

Would there be a way to approximate a solution if we assume W is positive and near zero but always greater than zero.

Because I can't solve it for all the solutions. Would there be a way to find the smallest positive solution.

 

[mfb]

Taylor series might work well. You can always find solutions to arbitrary precision with numerical methods. It depends on your problem then. Do you prefer an approximation that is not so good, but can be written down as formula, or an approximation that is much better, but needs dedicated calculation in each case?

 

[SSGD]

Yeah I just did both. Taylor Series gives me a number, but I was hoping to get several decimals of accuracy. But, the number of terms I will need is going to be... I think... Large.

The numerical solution gives the right askers, but It always wants to go to the Trivial Solution "0". With the constraints it worked.

Thanks for the help on this. I didn't think there was a way to get an general analytic solution. I would need to solve it for N's in the range of 20 to 100. So we are talking about some really large polynomials.

Again that's for the help.