How do I solve, tan^2(x)-(1+sqrt(3))tanx+sqrt(3)<0 ?

I know how to factor it, but then how would I get to the solutions? The domain is [pi/2,5pi/2].

1 Answer
Apr 11, 2018

color(blue)((pi/4,pi/3)uu((5pi)/4,(4pi)/3)uu((9pi)/4,(7pi)/3)

Explanation:

tan^2(x)-(1+sqrt(3))tan(x)+sqrt(3)<0

Let \ \ \ \ \u=tan(x)

u^2-(1+sqrt(3))u+sqrt(3)<0

Factor LHS:

(1-u)(-u+sqrt(3))<0

Solving for zero to get the bounds:

1-u=0=>u=1

-u+sqrt(3)=0=>u=sqrt(3)

But \ \ \ \ \u =tan(x)

tan(x)=1

tan(x)=sqrt(3)

x=arctan(tan(x))=arctan(1)=>x=pi/4, (5pi)/4, (9pi)/4

x=arctan(tan(x))=arctan(sqrt(3))=>x=pi/3, (4pi)/3, (7pi)/3

Using:

(1-u)(-u+sqrt(3))<0

and substituting u=tan(x)

(1-tan(x))(-tan(x)+sqrt(3))<0

For:

pi/4 < x < pi/3=3

(1-tan((5pi)/18))(-tan((5pi)/18))+sqrt(3))<0 \ \ \ true

For:

(5pi)/4 < x < (4pi)/3

(1-tan((23pi)/18))(-tan((23pi)/18))+sqrt(3))<0 \ \ \ true

For:

(9pi)/4 < x < (7pi)/3

(1-tan((55pi)/24))(-tan((55pi)/24))+sqrt(3))<0 \ \ \ true

Solutions:

color(blue)((pi/4,pi/3)uu((5pi)/4,(4pi)/3)uu((9pi)/4,(7pi)/3)