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

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve : arc cos (x - 1) = 2*arc cos x

take the cos of both the sides:

cos [arc cos (x - 1)] = cos [2*arc cos x]

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

=> x - 1 = 2*x^2 - 1

=> 2x^2  - x = 0

=> x(2x - 1) = 0

x = 0 and x = 1/2

The solution of the equation is x = 0 and x = 1/2

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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}.