find the general solution for :- tan²x + (1-√3) tanx - √3 = 0
Answers
Answer:
tan^2(x)-(1+sqrt(3))tan(x)+sqrt(3)<0
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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
Step-by-step explanation:
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