Gee whizz differential equations are fun!!
Hey Shadow, got another question for you. I have attached it in an image.
I am aware of the following rules:
If -1<a<1 and cosx = a, then the general solutions are x = 2n*pi +/- arccos(a), where n is any real integer
If -1<a<1 and tanx = a, then the general solutions are x = n*pi +/- arctan(a), where n is any real integer
If -1<a<1 and sinx = a, then the general solutions are x = 2n*pi + arcsin(a) or x = (2n+1)*pi - arcsin(a), where n is any real integer,
I can see from part (a) how you go from sinA = a, to write a solution for sinx = -a, as
x = arcsin(-a) = -arcsin(a) = -A
Thus solutions are x = 2*pi - A and x = pi + A
But when it comes to determining solutions when cosx = a, and we are given sinA = a to start, I am not so sure.
I thought you could use the compound angle formulae somehow, or substitute sinA for a such that
x = arcos(sinA),
Then taking the fact A is restricted to 0 < A < 1/2*pi, find the solution that way......but I am not sure as I have not seen this problem before, even in year 12. I was not much of a frontier math student in those days, and I would like to know how to do it.
You are right Shadow, Specialist Maths is a little difficult, but not if you know the basics and have some original thought behind your methods, like the above problem.
Thanks again,
James