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

1 Answer

May 9, 2015

Call tan x = t. Use the trig identity:# tan 2t = (2t)/(1 - t^2)#

#(2t)/(1 - t^2) = t -> 2t = t(1 - t^2) = t - t^3 -> t^3 + t = 0 #

#t(t^2 + 1) = 0 --> t = tan x = 0# -> x = pi and x = 2pi.

Check.
#x = pi -> tan x = 0 and tan 2x = tan (2pi) = 0# Correct.

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