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

1 Answer
Feb 17, 2017

#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#