Question #3c7f5

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Question #3c7f5

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Answer 1 · 2017-04-11T02:11:42

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.

Answer 2 · 2017-04-11T02:16:57

$x=pi/6$

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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