Solving Trig Equation tan^2x - 5tanx + 5 = 0 on (0, 2pi)

[mathcholo]

We are supposed to set the equation equal to zero to see what X equals within (0, 2pi)
I tried it two ways:

tan^2x-5tanx+5=0
(tanx-3)(tanx-2)=0
(tanx-3)=0 (tanx-2)=0
tanx=3 tanx=2

but then what? it doesn't seem to work out because in class the answer seemed to come out to be like tanx=sqrt. 3
then it the final answer would be pi/3 and 4pi/3.

My other method:
tan^2x-5tanx+5=0
tan^2x-5tanx=-5
tanx(tanx-5)=-5

but then I'm confused how i can set tanx=-5 or (tanx-5)=-5

 

[stapel]

We are supposed to set the equation equal to zero to see what X equals within (0, 2pi)
I tried it two ways:

tan^2x-5tanx+5=0
(tanx-3)(tanx-2)=0
(tanx-3)=0 (tanx-2)=0
tanx=3 tanx=2

but then what? it doesn't seem to work out because in class the answer seemed to come out to be like tanx=sqrt. 3

You might want to check with a fellow student. I really doubt that the instructor said that the solution to tan(x) = 3 or tan(x) = 2 was the square root of three! :shock:

then it the final answer would be pi/3 and 4pi/3.

When you plugged these values in for "x", did they work in the original equation, or did they fail? So maybe this is where you're supposed to apply what they covered when they taught about the "inverse" keys on your calculator.

My other method:
tan^2x-5tanx+5=0
tan^2x-5tanx=-5

When has this ever been an appropriate method for solving a quadratic equation? (Hint: Never.)

 

[Prove It]

Since when is (-3)(-2) = 5?

 

[Deleted member 4993]

We are supposed to set the equation equal to zero to see what X equals within (0, 2pi)
I tried it two ways:

tan^2x-5tanx+5=0
(tanx-3)(tanx-2)=0
(tanx-3)=0 (tanx-2)=0
tanx=3 tanx=2

but then what? it doesn't seem to work out because in class the answer seemed to come out to be like tanx=sqrt. 3
then it the final answer would be pi/3 and 4pi/3.

My other method:
tan^2x-5tanx+5=0
tan^2x-5tanx=-5
tanx(tanx-5)=-5

but then I'm confused how i can set tanx=-5 or (tanx-5)=-5

[h=2]tan^2x - 5tanx + 5 = 0 on (0, 2π)[/h]This is a quadratic equation of the form:

Ax^2 + Bx + C = 0

Then the solution for x is

x_1,2 = [-B ± √(B² - 4*A*C)]/[2*A]

Then

tan(x) = [5 ± √(25 - 4*1*5)]/2

Now leave it in fraction&radical form or get your calculator out and use it.....