Question

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

Answer 1

`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