How do you find all values of x in the interval [0, 2pi] in the equation #cos2x= cosx#?

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Answer 1 · 2016-03-09T14:26:58

#x=0, (2pi)/3, (4pi)/3, 2pi#

Using the identity #cos2x=2cos^2x-1#, #cos2x=cosx# becomes

#2cos^2x-1=cosx# or

#2cos^2x-cosx-1=0#

Using quadratic formula

#cosx=(-(-1)+-sqrt((-1)^2-4xx2xx(-1)))/(2xx2)# or

#cosx=(1+-sqrt(1+8))/4# or #cosx=(1+-3)/4# i.e.

#cosx=1 or -1/2#

Hence #x=0, (2pi)/3, (4pi)/3, 2pi#