How do you solve #sinxcotx=sqrt3/2#?

1 Answer
MathFact-orials.blogspot.com
Dec 10, 2016

#pi/6, (5pi)/6#

Explanation:

We have:

#sinxcotx=sqrt3/2#

#sinx(cosx/sinx)=sqrt3/2#

#cosx=sqrt3/2#

Cosine is #"adjacent"/"hypotenuse"=sqrt3/2#. This is the ratio of a 30-60-90 triangle (the missing length, the opposite, is length 1). So with the "short side" being the opposite, we know we're working with a reference angle of #30^o=pi/6#.

Since we have both the sine and cotangent functions in the original equation, we're looking for solutions where they are either both positive (Q1) or both negative (Q4). That means the angles are:

#pi/6, (5pi)/6#

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