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

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