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

Answer 1

Anna Andrews

#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}#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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