Question

How do you solve `3cot^2x-1=0` between the interval `0<=x<=2pi`?

Answer 1

In `(0, 2pi), x =pi/6, 5pi/6, 7pi/6 and 11pi/6`

Explanation:

Here, `tan^2 x=1/3`. So,

`tanx =+-1/sqrt3`.

The principal values for x are `+-pi/6` and the genera values are

`npi+-pi/6, n =0, +-1, +-2, +-3, ...`

`=+-pi/6, +-5pi/6, +-7pi/6 ...`

In `(0, 2pi), x =pi/6, 5pi/6, 7pi/6 and 11pi/6`.