Question

How do you use the quadratic formula to solve `4cos^2x-4cosx-1=0` in the interval `[0,2pi)`?

Answer 1

`101^@95 ; 258^@05`

Explanation:

Solve this quadratic equation for cos x by using the improved quadratic formula:
`f(x) = 4cos^2 x - 4cosx - 1 = 0`
`D = d^2 = b^2 - 4ac = 16 + 16 = 32` --> `d = +- 4sqrt2`
There are 2 real roots:
`cos x = - b/(2a) +- d/(2a) = 4/8 +- 4sqrt2/8 = (1 +- sqrt2)/2`
`cos x1 = (1 + sqrt2)/2 = 2.414/2 = 1.207` (Rejected as > 1)
`cos x2 = (1 - sqrt2)/2 = - 0.414/2 = - 0.207`
cos x2 = - 0.207 --> `x = +- 101^@95`
x = 258.05 is co-terminal to (- 101.95)
Answers for `(0, 2pi)`
`101^@95 ; 258^@05`