Question

Question #3c7f5

Answer 1

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

`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`