How do you solve #2cos2x=1-cosx# for #0<=x<=360#?

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Answer 1 · 2016-10-30T13:52:45

Apply the identity #cos2x = 1- 2sin^2x#.

#2(1 - 2sin^2x) = 1 - cosx#

#2 - 4sin^2x = 1 - cosx#

#2 - 4(1 - cos^2x) = 1 - cosx#

#2 - 4 + 4cos^2x = 1 - cosx#

#4cos^2x + cosx - 3 = 0#

Let #t = cosx#.

#4t^2 + t - 3 = 0#

#4t^2 + 4t - 3t - 3 = 0#

#4t(t + 1) - 3(t + 1) = 0#

#(4t - 3)(t + 1) =0#

#t = 3/4" AND " -1#

#cosx = 3/4" AND " cosx = -1#

#x = arccos(3/4), 360 - arccos(3/4)" AND "x = 180˚#

Hopefully this helps!