Question

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

Answer 1

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]