How do you solve #tan^2x-2tanx+1=0#?

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Answer 1 · 2016-11-06T23:53:40

Please see the explanation for steps leading to the answer:
#x = pi/4 + npi# where #n = ..., -3, -2, -1, 0, 1, 2, 3, ... #

Given:

#tan^2(x) - 2tan(x) + 1 = 0#

The above fits the pattern #a^2 + 2ab + b^2 = (a - b)^2# where #a = tan(x) and b = 1#:

#(tan(x) - 1)^2 = 0#

#tan(x) = 1#

#x = pi/4 + npi# where #n = ..., -3, -2, -1, 0, 1, 2, 3, ... #