Question

Solve the following equation on the interval (0,2π)

cos2x=cosx

Answer 1

x = 0, 2π/3, 4π/3, 2π.

Explanation:

cos 2x = 2cos^2x - 1 so

2cos^2x - 1 = cos x

2cos^2x - cos x - 1 = 0

(2cosx + 1)(cos x - 1) =0

cos x = -1/2, 1.

When cos x = 1, x = 0, 2π

when cos x = -1/2, x = 2π/3, 4π/3.