2 cos(x) = 1
Cosine is negative in 2nd & 3rd quadrant
--giving x as (2π/3) and (4π/3)
you were asked the general solution so the values become
(2π/3)+2nπ, (4π/3)+2nπ (as cos(2π+ y) = cos(y))
Even the title sounds scary =/. Well i'm given an equation like: 2cosx + 1 = 0 and it says to solve it. I know that the answer is: (2π/3)+2nπ, (4π/3)+2nπ. (That symbol after the 2 is the letter 'n' followed by the pi sign 'π') I have no idea whatsoever on how to do this. So pretty much any help is appreciated.
=)
Edit: And I just realized that there was a trigonometry section right after I posted this so hopefully it will get moved and sorry for the inconvenience.