Posted by **Constance** on Tuesday, November 6, 2012 at 9:35am.

2-sin^20/cos0 = sec0 + cos0

* 0 = Theta

- Math -
**Reiny**, Tuesday, November 6, 2012 at 10:05am
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

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