How do you find the general solutions for 6cos^2x - cos x - 1 = 0?

1 Answer
Aug 12, 2015

{ 2m pi +- pi/6, m in ZZ } uu {(2n+1) pi +- arccos(1/3), n in ZZ}

Explanation:

To simplify matters, take z = cos(x) , so the equation now becomes 6 z^2 - z - 1 = 0 .

This can be solved by factorisation, as follows.
6 z^2 - z - 1 = 0
=> 6 z^2 - 3 z + 2 z - 1 = 0
=> 3 z (2 z - 1) + (2 z - 1) = 0
=> (2 z - 1) (3 z + 1) = 0
=> z = 1/2 or z = -1/3

Now, replacing z = cos(x) , and solving for x, we get:

  • cos(x) = 1/2 => x = 2m pi +- pi/6, m in ZZ
  • cos(x) = -1/3
    => x = (2n) pi +- arccos(-1/3), n in ZZ
    => x = (2n+1) pi +- arccos(1/3), n in ZZ