Solve cos x -1 = cos 2x over the interval [0,2pi)

Please solve the equation and explain each step. Thank you.

cos x -1 = cos(2x)

=> cos(x) - 1 = 2cos^2(x) - 1

=> cos(x) - 2cos^2(x) = 0

=> cos(x) [ 1 - 2 cos(x) ] = 0

=> cos(x) = 0, 1 - 2cos(x) = 0

cos(x) = 0 and 1/2

x = π/2, 3π/2, π/3, 5π/3

Remember that cos2x = 2cos^2x - 1

Then,

cosx - 1 = 2cos^2x - 1

2cos^2x -cosx = 0

cosx (2cosx - 1) = 0

& you can continue now, to solve this simple equation...

You have to express cos2x as 2cos^2x - 1, this is a standard formula.
2cos^2x - cosx = 0
cosx(2cosx - 1) = 0
cosx = 0 or cosx = 1/2
x = pi/2 or 3pi/2, or pi/3 or 5pi/3

cos(x) - 1 = cos(2x)

cos(x) - 1 = 2*cos^2(x) - 1

cos^2(x) = 0

cos(x) = 0

x = pi/2, 3pi/2