Question

How do you solve `sqrt3+tan(2x)=0` between the interval `0<=x<=2pi`?

Answer 1

`pi/3; (5pi)/6`

Explanation:

`tan 2x = -sqrt3`
Trig table and unit circle gve 2 solutions:
`2x = (2pi/3)`, and `2x = (2pi/3) + pi = (5pi)/3`
a. `2x = (2pi)/3`
`x = pi/3`
b. `2x = (5pi)/3`
`x = (5pi)/6`
Check.
`x = pi/3` --> `tan 2x = tan ((2pi)/3) = -sqrt3`.
`sqrt3 + tan 2x = sqrt3 - sqrt3 = 0`. Proved
`x = (5pi)/6` --> `tan (10pi)/6 = tan (5pi)/3 = tan (2pi)/3 = - sqrt3`
`sqrt3 + tan 2x = sqrt3 - sqrt3 = 0`. Proved.