How do you find the general solutions for #cos(x/2) + cosx - 1 = 0#?

1 Answer
Nghi N.
Aug 13, 2015

Solve #f(x) = cos (x/2) + cos x - 1 = 0#

Ans: 77.48 + k360 deg

Explanation:

Call #cos (x/2) = t#
Use trig identity: #cos x = 2t^2 - 1.#
#f(x) = t + 2t^2 - 2 = 2t^2 + t - 2#.
#D = d^2 = b^2 - 4ac = 1 + 16 = 17# --> #d = +- sqrt17 = +- 4.12#

#t1 = -1/4 + sqrt17/4 = 3.12/4 = 0.78 # (accepted)
#t2 = -1/4 - sqrt17/4 = -5.12/4 = - 1.28 #(rejected)
#cos (x/2) = 0.78# --> #x/2 = 38/74# deg --> #x = 77.48# deg
General answer: x = 77.48 + k360.
Check.
x = 77.48 --> cos x = 0.22 --> x/2 = 38.74 --> cos (x/2) = 0.78.
cos (x/2) + cos x - 1 = 0.78 + 0.22 - 1 = 0. OK

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