solve over the interval [0, 2pi)
cos 2x - 3.cos x + 2 = 0
(2cos²x - 1) - 3cosx + 2 = 0
2cos² - 3cosx + 1 = 0
(2 cosx - 1).(cosx - 1) = 0
cosx = 1/2, cosx = 1
x = π/3, 5π/3, 0 in the interval given.
x=0
x = 0 or 2pi
Comments
cos 2x - 3.cos x + 2 = 0
(2cos²x - 1) - 3cosx + 2 = 0
2cos² - 3cosx + 1 = 0
(2 cosx - 1).(cosx - 1) = 0
cosx = 1/2, cosx = 1
x = π/3, 5π/3, 0 in the interval given.
x=0
x = 0 or 2pi