PRE CALC PLEASE HELP!!!!!!!!
posted by Hayli on .
find the solutions of the equation that are in the interval [0,2pi)
2cos^2x=4sin^2(1/2x)

2  (1  2sin^2 (x/2) = 4sin^2 (x/2)
1 = 2 sin^2 (x/2)
sin^2 (x/2) = 1/2
sin (x/2) = 1/√2
x/2 = π/4 or x/2 = 3π/4
then x = π/2 or x = 3π/2