Question

How do you solve `0=cos^2x-5cosx+1`?

Answer 1

`± arccos( (5 - sqrt 21)/2) + 2 pi \cdot ZZ`

Explanation:

`cos x = y` and `0 = y^2 -5y + 1`

Bhaskara. `Delta = 25 - 4`

`y = 5/2 ± sqrt 21/2`

With calculator, you see `y+ > 1` and `y- < 1`

`cos x = (5 - sqrt 21)/2 = cos a`

`x = a + 2k pi` or `x = - a + 2k pi`