Sunday
March 26, 2017

Post a New Question

Posted by on .

cos2x=cosx in interval of [0,2pie]

  • pre calc - ,

    cos2x-cosx=0
    expand the left hand side
    cos²(x)-sin²(x)-cos(x)=0
    2cos²(x)-1 - cos(x)=0
    substitute c=cos(x)
    2c²-c-1=0
    c=(-1±√(9))/4
    =-1 or 1/2
    cos(x)=-1 when x=π (0≤x≤2π)
    or
    cos(x)=1/2 when x=π/3 or x=5π/3 (0≤x≤2π)

    Substitute each of the three solution into the original equation to make sure that the solutions are acceptable.

  • pre calc - ,

    Reiny is right.
    There was a mistake in the solution of the quadratic.

    c=(1±√(9))/4
    =1 or -1/2
    cos(x)=1 when x=0 or 2π (0≤x≤2π)
    or
    cos(x)=-1/2 when x=π±π/3 (0≤x≤2π)

    Substitute each of the three solution into the original equation to make sure that the solutions are acceptable.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question