Math
posted by Constance .
2sin^20/cos0 = sec0 + cos0
* 0 = Theta

ok, so it is
2  sin^2 Ø/cosØ = secØ + cosØ ??
2  sin^2 Ø/cosØ = 1/cosØ + cosØ
times cosØ
2cosØ  sin^2 Ø = 1 + cos^2 Ø
2cosØ = 1 + sin^2 Ø + cos^2 Ø
2cosØ = 1+1
cosØ = 1
Ø = 0 , 2π or x = 0° , 360° , for the first period
general solution
x = 2kπ, where k is an integer