How do you find the general solutions for #tan^2 (x) = 1/3#?

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Answer 1 · 2015-08-22T23:45:19

Solve #tan^2 x = 1/3#

#tan^2 x = 1/3# --> #tan x = +- 1/sqrt3 = +- sqrt3/3#

a. #tan x = sqrt3/3# --> #x = pi/6 + kpi#

b. #tan x = - sqrt3/3 #--> #x = (5pi)/6 + kpi#

Answer 2 · 2015-08-22T23:55:32

#x=npi+-pi/6#

If #tan^2(x) = 1/3#

#rarr tan(x) = +-1/sqrt(3)#

The image below shows the possible configurations:
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