How to solve tg^2x-(1+sqrt3)tgx+sqrt3<0tg2x−(1+√3)tgx+√3<0?

1 Answer
Nghi N.
Jun 13, 2018

x = pi/3 + kpix=π3+kπ
x = pi/4 + kpix=π4+kπ

Explanation:

tan^2 x - (1 + sqrt3)tan x + sqrt3 = 0tan2x−(1+√3)tanx+√3=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 = 0.536D=d2=b2−4ac=(1+√3)2−4√3=4−2√3=0.536 --> d = +- 0.73d=±0.73
There are 2 real roots:
tan x = -b/(2a) +- d/(2a) = (1 + sqrt3)/2 +- 0.73/2 = 1.37 +- 0.37tanx=−b2a±d2a=1+√32±0.732=1.37±0.37
tan x = 1.37 + 0.37 = 1.73 = sqrt3tanx=1.37+0.37=1.73=√3, and
tan x = 1.37 - 0.37 = 1tanx=1.37−0.37=1
a. tan x = sqrt3tanx=√3
Trig table and unit circle give -->
x = pi/3 + kpix=π3+kπ
b. tan x = 1tanx=1
x = pi/4 + kpix=π4+kπ

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