Originally Posted by
brandonb122 Solve the equation over the intervals [0,2pi)
3cosx + 3 = 2sin²x
2sin²x - 3cosx - 3 = 0 Put it all on one side
2-2cos²x - 3cosx - 3 = 0 Used the pythagorean identity that Sin²x + Cos²x = 1
-2cos²x - 3cosx - 2 = 0 Combined the integers
From here i was confused because it doesn't factor into two binomials due to -cos²x being negative. So.. i made -cos²x positive using the even/odd identity. I am not sure if that is right.
(2cosx + 1)(cosx - 2) = 0 Used even/odd identity
So cosx = -1/2, 2
2pi/3, 4pi/3 My answer, didn't know what to do with the 2
Is this correct?