How do you solve #tan 3x(tanx-1)=0# in the interval 0 to 2pi?

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Answer 1 · 2016-02-13T08:18:14

Solution set is #{0, pi/4, pi/3, (2pi)/3, pi, (5pi)/4, (4pi)/3, (5pi)/3, 2pi}#

As #tan3x(tanx−1)=0#, this means

Either #tan3x=0# i.e.

#3x=0 or pi or 2pi or 3pi or 4pi or 5pi or 6pi#

i.e. #x=0 or pi/3 or (2pi)/3 or pi or (4pi)/3 or (5pi)/3 or 2pi#

Alternatively #(tanx−1)=0# i.e. #tanx=1#

i.e. #x=pi/4 or (5pi)/4#

Hence Solution set is #{0, pi/4, pi/3, (2pi)/3, pi, (5pi)/4, (4pi)/3, (5pi)/3, 2pi}#