Question #3c7f5

2 Answers
Cri007
Apr 11, 2017

See below.

Explanation:

We move all the tangents to one side.

#3=3sqrt(3)tanx#
So #tanx=1/sqrt(3)#

This happens at #pi/6#, and since tangent's period is #pi#, this occurs:

#pi/6+kpi#, where #k# is an integer.

Rachel
Apr 11, 2017

#x=pi/6#

Explanation:

I'm not sure if I'm answering your question, but here goes:

We can think of #tanx#, for now at least, as a variable. For that reason, we'll let #tanx=u#

#5sqrt(3)u+3=8sqrt(3)u#
Subtract #5sqrt(3)u# on both sides
#3=3sqrt(3)u#
Divide by #3sqrt(3)# on both sides
#cancel(3)/(cancel(3)sqrt(3))=u#
If we rationalize the denominator we get #sqrt(3)/3=u# or #sqrt(3)/3=tanx#

To solve for #x#, apply the arc tangent (#tan^(-1)#).

#tan^(-1)(sqrt(3)/3)=x#, which is #pi/6#

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