*to find all solutions in the interval [0,2pi):*

**quadratic formula**This is for my trig class. Thanks for the help!

Tutors, sign in to answer this question.

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

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.

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.

## Comments