Question

How do you solve `3tan^2theta=1`?

Answer 1

The solutions are `S={pi/6+kpi , -pi/6+kpi}`

Explanation:

`3tan^2theta=1`

`tan^2theta=1/3`

`tantheta=+-1/sqrt3`

`tantheta=1/sqrt3`

`theta=pi/6 +kpi`, k in ZZ##

`tan theta=-1/sqrt3`

`theta=-pi/6+kpi`, `kinZZ`

Answer 2

General solution: `theta = 2npi + (pi/6) , 2n pi+( (7pi)/6) , 2npi + ((5pi)/6) , 2n pi +( (11pi)/6)` for `[n=0,1,2,3, .........]`

Explanation:

`3 tan^2 theta = 1 or tan^2 theta = 1/3 or tan theta = +- 1/sqrt 3, tan theta = 1/sqrt 3 ; tan (pi/6) = 1/sqrt 3 ; tan (pi +pi/6) = 1/sqrt 3 :. theta = (pi/6) , theta = ( (7pi)/6)`

`tan theta = -1/sqrt 3 ; tan (pi -pi/6) = -1/sqrt 3 ; tan ( 2pi -pi/6) = -1/sqrt 3 :. theta = ((5pi)/6) , theta = ( (11pi)/6)`

General solution: `theta = 2npi + (pi/6) , theta = 2n pi+( (7pi)/6)`

and `theta = 2npi + ((5pi)/6) , theta = 2n pi +( (11pi)/6)`

where `n` is a whole number as `[n=0,1,2,3, .........]` [Ans]