How do you use the quadratic formula to solve #3tan^2x+3tanx-4=0# in the interval #[0,2pi)#?

1 Answer
Nov 16, 2017

#37^@17; 119^@60; 217^@17; 299^@60#

Explanation:

#f(x) = 3tan^2 x + 3tan x - 4 = 0#.
Solve this quadratic equation for tan x by using the improved formula (Socratic, Google Search):
#D = d^2 = b^2 - 4ac = 9 + 48 = 57# --> #d = +- sqrt57 = +- 7.55#
There are 2 real roots:
#tan x = -b/(2a) +- d/(2a) = - 3/6 +- 7.55/6 = (-3 +- 7.55)/6#
#tan x = - 10.55/6 = - 1.76#
#tan x = 4.55/6 = 0.76#
a. tan x = - 1.76
Calculator and unit circle give:
#x = - 60^@40# and #x = -60.40 + 180 = 119^@60#
The arc (#-60^@40#) is co-terminal to arc
(#360 - 60.40 = 299^@60#)
b. tan x = 0.76
#x = 37^@17# and #x = 37.17 + 180 = 217^@17#
Answers for (0, 360):
#119^@60; 299^@60; 37^@17; 217^@17