Question

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

Answer 1

`{ 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`