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

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

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
or

Quadratic formula is driven by , factoring by completing square of general quadratic of ax^2 + bx +c

aX^2 + bX + C =0

X ^2 + b/a X + C/a = 0

X^2 + 2 b/2a  =   -c/a

X^2 +2 b/2a + b^2/(4a^2) = -c/a +  b^2/ (4 a^2)

( X + b/ 2a) ^2  = ( b^2 - 4ac) / 4a^2

X + b/2a = ±√(b^2- 4ac)/ 2a

X = -b/2a ±√(b^2 -4ac) / 2a