Question

How do you find all solutions of the equation in the interval `[0,2pi)` given `sqrt3tan3x=0`?

Answer 1

`x={0,pi/3,(2pi)/3,pi,(4pi)/3,(5pi)/3}`

Explanation:

`sqrt2tan3x=0=>tan3x=0` as `sqrt3!=0`

Hence `3x=npi` or `x=(npi)/3`

and in the interval `[0,2pi)`

`x={0,pi/3,(2pi)/3,pi,(4pi)/3,(5pi)/3}`