How to solve #tg^2x-(1+sqrt3)tgx+sqrt3<0#?

1 Answer
Nghi N.
Jun 13, 2018

#x = pi/3 + kpi#
#x = pi/4 + kpi#

Explanation:

#tan^2 x - (1 + sqrt3)tan x + sqrt3 = 0#
Solve this quadratic equation for tan x.
Use the improved quadratic formula (Socratic, Google Search)
#D = d^2 = b^2 - 4ac = (1 + sqrt3)^2 - 4sqrt3 = 4 - 2sqrt3 = 0.536# --> #d = +- 0.73#
There are 2 real roots:
#tan x = -b/(2a) +- d/(2a) = (1 + sqrt3)/2 +- 0.73/2 = 1.37 +- 0.37#
#tan x = 1.37 + 0.37 = 1.73 = sqrt3#, and
#tan x = 1.37 - 0.37 = 1#
a. #tan x = sqrt3#
Trig table and unit circle give -->
#x = pi/3 + kpi#
b. #tan x = 1#
#x = pi/4 + kpi#

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