How do you solve for the exact solutions in the interval [0,2pi] of #cos 2x - cos x= 0#?

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Answer 1 · 2016-03-14T05:10:37

Solution set within #[0,2pi]# is #{0, (2pi)/3, (4pi)/3, 2pi}#

As #cos2x=2cos^2x-1#, the equation becomes

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

and using quadratic formula

i.e. #cosx=(-(-1)+-sqrt((-1)^2-4xx2xx(-1)))/(2xx2)#

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

i.e. #cosx=1 or -1/2#

As #cosx=1# for #x=0 or 2pi#

and #cosx=-1/2# for #x=(2pi)/3 or (4pi)/3#

Solution set within #[0,2pi]# is #{0, (2pi)/3, (4pi)/3, 2pi}#