Solve the equation arccos(x-1) = 2arccosx

First, we'll impose the constraints of existence of the
function
arccosine.

|x-1|=<1

|x|=<1

We'll
apply cosine function both sides:

cos [arccos(x-1)] = cos
(2 arccos x)

By definition, cos (arccos a) = a =>
cos [arccos(x-1)] = x - 1

Also cos (2 arccos x) = 2 [cos
(arccos x)]^2 - 1

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

The equation will become:

x
- 1 = 2x^2 - 1

2x^2 - x =
0

x(2x - 1) = 0

We'll cancel
each factor:

x = 0

2x - 1 =
0

x = 1/2

Since
both values of x respect the contraints of existence, then the solutions of the
equations are: {0 , 1/2}.