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

Question

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

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Answer 1 · 2018-04-11T15:13:24

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

#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)#