How do you solve the following equation # tan^2 x = 1/3 # in the interval [0, 2pi]?

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How do you solve the following equation # tan^2 x = 1/3 # in the interval [0, 2pi]?

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Answer 1 · 2016-10-31T13:19:30

#pi/6, (5pi)/6, (7pi)/6, (11pi)/6#

Explanation:

#tan^2 x = 1/3#
#tan x = +- 1/sqrt3 = +- sqrt3/3#
Use unit circle and trig table -->
a. #tan x = sqrt3/3# --> #x = pi/6#, and #x = pi/6 + pi = (7pi)/6#
b. #tan x = -sqrt3/3# --> #x = (5pi)/6# and #x = (5pi)/6 + pi = (11pi)/6#

Answers for #(0, 2pi)#:
#pi/6, (5pi)/6, (7pi)/6, (11pi)/6#

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How do you solve the following equation # tan^2 x = 1/3 # in the interval [0, 2pi]?

1 Answer

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