How do you solve #\tan ^ { 2} x - 4\tan x + 1= 0#?

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Answer 1 · 2018-02-09T22:52:00

#color(blue)((5pi)/12+npi)#

#color(blue)(pi/12+npi)#

#tan^2x-4tanx+1=0#

Let #color(white)(88)u=tanx#

Then:

#u^2-4u+1=0#

Using Quadratic Formula:

#u=(-(-4)+-sqrt((-4)^2-(4*1*1)))/(2*1)#

#u=(4+-sqrt(12))/2=2+-sqrt(3)#

#u=tanx=2+-sqrt(3)#

#x=arctan(2+sqrt(3))=color(blue)((5pi)/12+npi)#

#x=arctan(2-sqrt(3))=color(blue)(pi/12+npi)#

For: #{n in ZZ}#