1 + cosy/2+ cos(2y)/4 + cos(3y)/8 ......... =(4-2cosy)/(5-4cosy)

this can be done using infinite geometric series, and taking the real part.

Then cos(4y) =8({(cosy)^4} -{(cosy)^2}) + 1

which can be done using double angle formulae.

Now we need to solve

x^4 - x^2 + 1/16 = 0

This can be done easily using complex no's, but I'm not sure how to do it using the previous results. I've tried setting cos(4y) = 1/16, and then letting x=cosy, but I end up with some horrible expression involving cos(0.25*arcos(1/16)) which is obviously not what I'm supposed to get.

Thanks