How do you solve #tan(2u) = tan(u)# on the interval [0,2pi)?

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Answer 1 · 2016-06-23T09:23:53

#u={0,pi}#

As #tan2u=(2tanu)/(1-tan^2u)#, #tan2u=tanu# can be written as

#(2tanu)/(1-tan^2u)=tanu#

or #2tanu=tanu(1-tan^2u)=tanu-tan^3u#

or #tan^3u+2tanu-tanu=0#

or #tan^3u+tanu=0#

or #tanu(tan^2u+1)=0#

As #(tan^2u+1)# is always positive and non-zero

#tanu=0# and #u={0,pi}#