How do you find all solutions in [0, 2π): #3tan^2x - 1 = 0#?

Question

How do you find all solutions in [0, 2π): #3tan^2x - 1 = 0#?

1 Answer

Impact of this question

4600 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-11T10:30:55

Solution: In #[0,2pi) x=pi/6,x=(7pi)/6 and x=(5pi)/6, x= (11pi)/6#

Explanation:

#3tan^2x-1=0 or3tan^2x=1 or tan^2x=1/3 or tanx=+- 1/sqrt3#. We know #tan(pi/6)= 1/sqrt3 and tan(pi+pi/6)= 1/sqrt3 # and also
#tan(2pi-pi/6)= -1/sqrt3 and tan(pi-pi/6)= -1/sqrt3 #.
Solution: In #[0,2pi) x=pi/6,x=(7pi)/6 and x=(5pi)/6, x= (11pi)/6#[Ans]

Science

Math

Humanities

... and beyond

How do you find all solutions in [0, 2π): #3tan^2x - 1 = 0#?

1 Answer

Explanation:

Related questions

Impact of this question