How do you solve cotx-sqrt3=0?

1 Answer
MathFact-orials.blogspot.com
Oct 28, 2016

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

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
8668 views around the world
You can reuse this answer
Creative Commons License