How do you find solutions of the equation in the interval $[0, 2 π]$ $[0,2pi]$ for $tan 2 x = tan x$ $tan2x=tanx$ ?

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Answer 1 · 2015-05-09T22:58:32

Call tan x = t. Use the trig identity: $tan 2 t = \frac{2 t}{1 - t^{2}}$ $tan 2t = (2t)/(1 - t^2)$

$\frac{2 t}{1 - t^{2}} = t \to 2 t = t (1 - t^{2}) = t - t^{3} \to t^{3} + t = 0$ $(2t)/(1 - t^2) = t -> 2t = t(1 - t^2) = t - t^3 -> t^3 + t = 0$

$t (t^{2} + 1) = 0 - \to t = tan x = 0$ $t(t^2 + 1) = 0 --> t = tan x = 0$ -> x = pi and x = 2pi.

Check.
$x = π \to tan x = 0 and tan 2 x = tan (2 π) = 0$ $x = pi -> tan x = 0 and tan 2x = tan (2pi) = 0$ Correct.

How do you find solutions of the equation in the interval [0,2pi][0,2π][0,2pi] for tan2x=tanxtan2x=tanxtan2x=tanx?