What are the solutions to 10(sec^2)x+5(tan^2)x-15=0 over the interval (0,2pi)?

1 Answer
Nghi N
May 11, 2018

#pi/6; (2pi)/3; (7pi)/6; (5pi)/3#

Explanation:

#10sec^2 x + 5tan^2 x - 15 = 0#
Replace #sec^2 x# by #(1 + tan^2 x)# from trig identity.
#10(1 + tan^2 x) + 5tan^2 x - 15 = 0#
#10tan^2 x + 5tan^2 x - 5 = 0#
#15tan^2 x = 5# --> #tan^2 x = 1/3#
#tan x = +- sqrt3/3#
Trig table and unit circle give 4 solutions for tan x -->
a. #tan x = sqrt3/3# -->
#x = pi/6# and #x = pi/6 + pi = (7pi)/6#
b. #tan x = - sqrt3/3# -->
#x = (2pi)/3# and #x = pi + (2pi)/3 = (5pi)/3#

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