Question

How do you find the solution to `tan^2theta-3=0` if `0<=theta<360`?

Answer 1

`theta=pi/3, (4pi)/3`

Explanation:

We have

`tan^2theta-3=0`

We can solve for `theta`:

`tan^2theta=3`

`tantheta=sqrt3`

We can express `tan=sin/(cos)=sqrt3/1`

The triangle that has the ratio of `1, sqrt3,2` is the 30-60-90 triangle. We have the opposite side as the medium length side, so we're talking about a reference angle of `60^o=pi/3`

Next thing to ask is where is tangent positive? Q1, Q3.

In Q1, we have `pi/3`

In Q3, we have `pi+pi/3=(4pi)/3`