Question

How do you solve `cos^2 x - 4 cos x = 1` over the interval 0 to 2pi?

Answer 1

`103^@65 and 256^@35`

Explanation:

`f(x) = cos^2 x - 4cos x - 1 = 0`
`D = d^2 = b^2 - 4ac = 16 + 4 = 20` ---> `d = +- 2sqrt5`
There are 2 real roots.
`cos x = -b/(2a) +- d/(2a) = 4/2 +- 2sqrt5/2 = 2 +- sqrt5`

a . `cos x = 2 + sqrt5` (Rejected because > 1)
b. `cos x = 2 - sqrt5 = 2 - 2.24 = - 0. 24` -->`x = +- 103^@65`
Answer for `(0, 2pi):`
`103^@65 and 256^@35` (co-terminal to - 103.65)