Question

How do you solve `cos2x=2cosx-1` in the interval [0, 2pi]?

Answer 1

`0, pi/2, 3pi/2 and 2pi`.

Explanation:

`cos 2x = 2 cos^2x-1`.

The equation reduces to

`cos x (cosx - 1) = 0`.

cos x = 0 `to x = pi/2 and 3pi/2, x in [0, 2pi]`.

cos x = 1 `to x=0 and 2pi, x in [0, 2pi]`.