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How do you solve #2 + 3cos2x = cosx#?

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May 13, 2015

Replace cos 2x by #(2cos^2 x - 1)#
f(x) = 2 + 3(2cos^2 x - 1) - cos x = 0. Call cos x = t.

6t^2 - t - 1 = 0. Solve by the formula
#D = d^2 = 1 + 24 = 25 -> d = +- 5#

#t = 1/12 + 5/12 = 6/12 = 1/2#

#t = 1/12 - 5/12 = -4/12 = -1/3#
Use the trig unit circle as proof:

#cos x = t = 1/2 -> x = pi/3 and x = 5pi/3#

#cos x = -1/3 -> x = 109.47 and x = 250.53 deg#

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