Question

How do you solve `cotx-sqrt3=0`?

Answer 1

`x=pi/6+n2pi, (7pi)/6+n2pi`, where n is an integer

Explanation:

Start with:

`cotx-sqrt3=0`

`cotx=sqrt3`

This means that:

`cosx/sinx=sqrt3/1`

So the adjacent is length `sqrt3` and the opposite is length 1. This set us up for the 30/60/90 triangle and the angle we're looking at is the `30^o` or `pi/6` one.

So now the question is which quadrants are we looking at. Cotangent is positive in Q1 and Q3. The angles we're looking at, then, are `pi/6` and `pi+pi/6=(7pi)/6`.

This question does not limit our answer to any particular domain, and so we need the general solution, which is:

`x=pi/6+n2pi, (7pi)/6+n2pi`, where n is an integer