f(x) = cos 2x - cos x - 1 = 0
Replace cos 2x by (2cos^2 x - 1)(2cos2x−1) -->
f(x) = 2cos^2 x - cos x - 2 = 0f(x)=2cos2x−cosx−2=0
Solve this quadratic equation for cos x -->
D = b^2 - 4ac = 1 + 16 = 17D=b2−4ac=1+16=17 --> d = +- sqrt17d=±√17
There are 2 real roots:
cos x = -b/(2a) +- d/(2a) = 1/4 +- sqrt17/4 = (1 +- sqrt17)/4cosx=−b2a±d2a=14±√174=1±√174
a. cos x = (1 + 4.12)/4 = 5.12/4 = 1.28cosx=1+4.124=5.124=1.28 (rejected because > 1)
b. cos x = (1 - 4.12)/4 = - 3.12/4 = - 0.78cosx=1−4.124=−3.124=−0.78
Calculator and unit circle give -->
x = +- 141^@26 = k360^@x=±141∘26=k360∘
Check by calculator.
x = 141^@26x=141∘26 --> cos x = - 0.78 --> 2x = 282^@522x=282∘52 --> cos 2x = 0.22 -->
f(x) = 0.22 - (- 0.78) - 1 = 0. Proved.
