How do you solve #cos 2(x)- cos(x)- 1 = 0#?

1 Answer
Nghi N
Jun 25, 2017

#x = +- 141^@26 + k360^@#

Explanation:

f(x) = cos 2x - cos x - 1 = 0
Replace cos 2x by #(2cos^2 x - 1)# -->
# f(x) = 2cos^2 x - cos x - 2 = 0#
Solve this quadratic equation for cos x -->
#D = b^2 - 4ac = 1 + 16 = 17# --> #d = +- sqrt17#
There are 2 real roots:
#cos x = -b/(2a) +- d/(2a) = 1/4 +- sqrt17/4 = (1 +- sqrt17)/4#
a. #cos x = (1 + 4.12)/4 = 5.12/4 = 1.28# (rejected because > 1)
b. #cos x = (1 - 4.12)/4 = - 3.12/4 = - 0.78#
Calculator and unit circle give -->
#x = +- 141^@26 = k360^@#
Check by calculator.
#x = 141^@26# --> cos x = - 0.78 --> #2x = 282^@52# --> cos 2x = 0.22 -->
f(x) = 0.22 - (- 0.78) - 1 = 0. Proved.

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