Posted by Hayli on Thursday, November 4, 2010 at 11:15pm.
find the solutions of the equation that are in the interval [0,2pi)
2cos^2x=4sin^2(1/2x)

PRE CALC PLEASE HELP!!!!!!!!  Reiny, Friday, November 5, 2010 at 8:40am
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