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 $u={0,pi}$

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