Solve this: #abs(2cos3x)=1# ---> (x=... + ...)?

Question

#abs(2cos3x)=1#

Will it be: #x=pi/9+2/3kpi# AND #x=-pi/9+2/3kpi#
OR ......
#x=pi/9+1/3kpi# AND #x=-pi/9+1/3kpi#

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Answer 1 · 2017-04-02T15:54:45

#x=2/3kpi+-pi/9# and #x=2/3kpi+-(2pi)/9#

As #|2cos3x|=1#, we have

either #2cos3x=1# i.e. #cos3x=1/2=cos(pi/3)#

and #3x=2kpi+-pi/3# or #x=2/3kpi+-pi/9#

or #2cos3x=-1# i.e. #cos3x=-1/2=cos((2pi)/3)#

and #3x=2kpi+-(2pi)/3# or #x=2/3kpi+-(2pi)/9#