It’s easy to check that…
Well then I’m glad I was able to do that much at least. Have a good day.
I assume you mean that it was originally 2sinc - sin2c, or else it would just be sinc.
Either way I have consulted with a friend regarding this equation and apart from reducing it to a quartic and then either using numerical methods or Ferraria’s Formula, we can’t think of another way to solve this question.
Sorry for not being of much help, but with these methods you will be able to solve the equation, Newton-Rhapson is the recommended approach as Ferraria’s Formula takes an insanely long time to use by hand.
Sorry I couldn’t be of more help to you, but if I think of anything or if anyone tells me how to do it today I’ll post it.
I’m sorry but as far as I can tell I have been largely unsuccessful in solving the equation you have given me. I have however been able to reduce everything down to a quartic which can be solved, it is just incredibly tedious to do by hand and most definitely should be assisted by some program or calculator.
I turned it into a quartic by squaring both sides and changing (sinc)^2 into 1-(cosc)^2 and then multiplying out. However, due to the weird fraction on the right hand side of the equation, I was unable to find an easy factorisation and instead found solutions online, which you can then find the arccos of in order to get values for c. If this is how you choose you to it you must remember to check the values satisfy the original equation, since you’ve squared both sides which means some solutions won’t be solutions to the original equation.
Besides that, I have tried using compound angle formulae and the factor formulae, but I have been so far unable to find a better way to solve the equation. I will however keep researching and if I find I method of solving your equation that is better than what I have already suggested then I will post it.
I’m sorry that I have been unable to adequately solve your equation, but I hope that what I managed to do has helped.
Hey, sorry but I was wrong when I said there were no roots, after graphing the function it becomes clear that there are two roots. However, as I am heading into school once more I don’t have time o find an algebraic way of factorising or deducing these two roots.
There are however methods covered on the Wikipedia page for Quartic Equations that detail how to solve them. Alternatively there are polynomial solvers online that you can use to solve quartics.
I will work on this as soon as I get home and If I find a way to do it I will post it.
Regarding the left hand bracket, what I meant was the bracket on the left hand side of the equals sign, so in this case I was referring to (1+c^2)^2
I am currently on my way to school so I might not be able to go into this too much but if it is real roots you are looking for then I don’t think there are any and this is why.
Diving by 2 you get (1+c^2)^2 = 5c
Now the bracket on the left hand side will always be non-negative since it is a squared bracket, and in fact the lowest point at (0,1). If you sub in c=0 to the right hand side you get the point (0,0) which is therefore below the curve.
If you then imagine sketching the rest of it out you’ll see that there isn’t an intersection with the real axes and so there is no real solution I don’t think.
If however you wanted complex roots then please let me know and I’ll work on it as soon as I get home.
Hope this helped.
Update: after finding a graphics calculator and drawing the graph there do seem to be two roots to the equation. As for how they are deduced from the equations, I’m tempted to say by using formulas often employed when solving quartics which is covered on Wikipedia.
One reason why mathematics enjoys special esteem, above all other sciences, is that its laws are absolutely certain and indisputable, while those of other sciences are to some extent debatable and in constant danger of being overthrown by newly discovered facts.