Find all solutions to the equation 8sin^2(theta) - 4 = 0 in the interval 2(pi) < (theta) < 4(pi)

First substiute sin(theta) = x (This step is not nesecessary but often people find quadratics easier to solve in variables that they usually use). Then due to the fact that the "b" coefficient in this equation is 0 and 4 is a perfect square, factorisation is simply a case of difference to two squares. Dividing through by 2 and factorising we get (2x - sqrt(2)(2x + sqrt(2) = 0. This gives us x = ±sqrt(2)/2.

Resubsitite sin(theta) = x and using basic trig we find the 4 solutions in the interval 0 < (theta) < 2(pi). The question asks about an interval different to this however, but the interval is a whole cycle over from the interval in which we calculated so we can simply add 2(pi) to each solution. Hence the answer is (theta) = 9(pi)/4, 11(pi)/4 for the positive root 2 over 2, and 13(pi)/4, 15(pi)/4 for the negitive root 2 over 2.