Use the quadratic formula to find all solutions in the interval [0,2pi): tan^2x-6tanx+4=0

Question

This is for my trig class. Thanks for the help!

Answer 1

tan^2(x) -6tanx + 4 = 0

let y = tanx, then

y^2 -6y +4 = 0

complete the square

y^2 -6x +9 = -4+9 = 5

(y-3)^2 = 5

y-3 = +/-sqr5

y = tanx = 3+/-sqr5

x = arctan(3+/-sqr5) = about 79.19 or 37.38 degrees

+ 180n where n = any integer

angle has to be in quadrant 1 or 3 where tangents are positive

Answer 2

Use quadratic formula to find all solutions in the interval [0,2pi):

tan^2(x) - 6 tan(x) + 4=0

 

tan(x) = (--6 ± sqrt((-6)^2 - 4(1)(4)))/(2(1))

= 3 ± sqrt( (2*3)^2 - 4(2^2) ) / (sqrt(2^2))

= 3 ± sqrt( (3)^2 - 4 )

= 3 ± sqrt( 5 ) ≈ 0.76393202250021 or 5.23606797749979

 

We can use calculator's tan^(-1)(x) function to find 1st quadrant solutions and add pi to both to get 3rd quadrant solutions (in radians):

 

x ≈ tan^(-1)(0.76393202250021) ≈ 0.652358139784368 or 3.79395079337416

x ≈ tan^(-1)(5.23606797749979) ≈ 1.382085796011335 or 4.52367844960113

 

pi * 0.5 sec / (60 sec/min) / (60 min/deg) / 180° ≈ 0.00000242406841 radians

 

So if we want our answers accurate to the nearest minute of arc we should round to the 5th decimal place:

 

x ≈ 0.65236 or 1.38209 or 3.79395 or 4.52368 radians

Answer 3

Completing the square,

(tanx-3)^2 = 9-4 = 5

tan x = 3+sqrt(5), x = 1.38 rad, 4.52 rad

or

tan x = 3-sqrt(5), x = .65 rad, 3.79 rad.

Answer: x = .65 rad, 1.38 rad, 3.79 rad, 4.52 rad.