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

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2 Answers

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.

Comments

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 

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