SOLUTION: The solutions of cos2x � cos x � 2 = 0 are (where k denotes an arbitrary integer) I know the answer is −π + 2kπ, I just have no idea how to get to that point.

Question 384531: The solutions of cos2x � cos x � 2 = 0 are (where k denotes an arbitrary integer)
I know the answer is −π + 2kπ, I just have no idea how to get to that point.
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
If you let z = cos x, then the equation becomes a simple quadratic:
(I hope the cos2x means , otherwise we'd need the power reduction formulas).
From the quadratic, we can factor and obtain --> z = 2 or z = -1. However, z is only defined on [-1, 1] since it is the range of cosine, so z = -1, and x = -pi (plus any multiple of 2pi since they denote the same angle on a unit circle).

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SOLUTION: The solutions of cos2x � cos x � 2 = 0 are (where k denotes an arbitrary integer) I know the answer is &#8722;&#960; + 2k&#960;, I just have no idea how to get to that point.

SOLUTION: The solutions of cos2x � cos x � 2 = 0 are (where k denotes an arbitrary integer) I know the answer is −π + 2kπ, I just have no idea how to get to that point.