Question #3c7f5

Question

Question #3c7f5

2 Answers

Impact of this question

1161 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-11T02:11:42

See below.

Explanation:

We move all the tangents to one side.

$3 = 3 \sqrt{3} tan x$ $3=3sqrt(3)tanx$
So $tan x = \frac{1}{\sqrt{3}}$ $tanx=1/sqrt(3)$

This happens at $\frac{π}{6}$ $pi/6$ , and since tangent's period is $π$ $pi$ , this occurs:

$\frac{π}{6} + k π$ $pi/6+kpi$ , where $k$ $k$ is an integer.

Answer 2 · 2017-04-11T02:16:57

$x = \frac{π}{6}$ $x=pi/6$

Explanation:

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

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

$5 \sqrt{3} u + 3 = 8 \sqrt{3} u$ $5sqrt(3)u+3=8sqrt(3)u$
Subtract $5 \sqrt{3} u$ $5sqrt(3)u$ on both sides
$3 = 3 \sqrt{3} u$ $3=3sqrt(3)u$
Divide by $3 \sqrt{3}$ $3sqrt(3)$ 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 $pi/6$

Science

Math

Humanities

... and beyond

Question #3c7f5

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question