SOLUTION: Résoudre dans R : Arccos (2x) - Arccos(x) = π/3

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Résoudre dans R : Arccos (2x) - Arccos(x) = π/3      Log On


   



Question 1149410: Résoudre dans R : Arccos (2x) - Arccos(x) = π/3
Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.
Résoudre dans R : Arccos (2x) - Arccos(x) = π/3
~~~~~~~~~~~~~~~~~~~



Arccos (2x)  -  Arccos(x) = π/3   ===========>


arccos(2x) = pi%2F3 + arccos(x).


Apply cos to both sides.


cos(acrcos(2x)) = cos%28pi%2F3%29%2Acos%28arccos%28x%29%29 - sin%28pi%2F3%29%2Asin%28arccos%28x%29%29.    (1)


Now, part by part,


    cos(arccos(2x)) = 2x;

    cos(pi/3)       = 1%2F2;

    cos(arccos(x))  = x;

    sin%28pi%2F3%29 = sqrt%283%29%2F2;

    sin(arccos(x)) = sqrt%281-x%5E2%29.


Substitute everything into equation (1).


    2x = %281%2F2%29%2Ax - %28sqrt%283%29%2F2%29%2Asqrt%281-x%5E2%29


    4x = x             - sqrt%283%29%2Asqrt%281-x%5E2%29

    3x = - sqrt%283%29%2Asqrt%281-x%5E2%29


Square both sides


    9x*2 = 3*(1-x^2)

    9x^2 = 3 - 3x^2

    12x^2 = 3

       x^2 = 3%2F12 = 1%2F4

       x   = +/- sqrt%281%2F4%29 = +/- 1%2F2.


Now the last step is to check these two candidates.


CHECK.


Case 1.  x = 1%2F2.

                Then 2x = 1,  arccos(2x) = arccos(1) = 0  radians.

                arccos(x) = arccos%281%2F2%29 = pi%2F3.

                0 - pi%2F3 = - pi%2F3.

                The original equation IS NOT satisfied, so  x = 1%2F2  IS NOT the solution.



Case 2.  x = - 1%2F2.

                Then 2x = -1,  arccos(2x) = arccos(-1) = pi  radians.

                arccos(x) = arccos%28-1%2F2%29 = 2pi%2F3.

                pi - 2pi%2F3 = pi%2F3.

                The original equation IS satisfied, so  x = - 1%2F2  IS  the solution.


ANSWER.  The only solution to the given equation is  x = - 1%2F2.

Solved.

-----------------

Come again to the forum soon to learn something new (!)