Question

How do you graph to solve the equation on the interval `[-2pi,2pi]` for `tanx=sqrt3`?

Answer 1

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

Explanation:

Since tangent is positive, we know that the angle comes from either QI or QIII.

Since `tan(x) = sqrt(3)` we know that `x` is a `pi/3` angle.

Within `0<x<2pi`, we know that the angles `pi/3` and `4pi/3` have tangents of `sqrt(3)`.

Since we're solving on `-2pi<x<2pi`, we can subtract `2pi` from each value and stay in the necessary interval:

`pi/3-2pi=-(5pi)/3`
`4pi/3-2pi=-(2pi)/3`

So our answers are `x={-(5pi)/3, -(2pi)/3, pi/3, (4pi)/3}`.