Question

High school math question

Answer 1

Answer: Choice C.
`x = \pm(\pi)/(6)`

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Step-by-step explanation:

Recall that

`\tan(x) = (\sin(x))/(\cos(x))\\\\`

So we can say

`\tan(3x) = (\sin(3x))/(\cos(3x))\\\\`

The vertical asymptotes occur when the denominator is zero.

If we were to plug in
`x = (\pi)/(6)`, then we'd have,

`\tan(3x) = (\sin(3x))/(\cos(3x))\\\\\tan(3*(\pi)/(6)) = (\sin(3*(\pi)/(6)))/(\cos(3*(\pi)/(6)))\\\\\tan((\pi)/(2)) = (\sin((\pi)/(2)))/(\cos((\pi)/(2)))\\\\\tan((\pi)/(2)) = (1)/(0)\\\\\tan((\pi)/(2)) = \text{und}\text{efined}\\\\`

This shows that one vertical asymptote is at
`x = (\pi)/(6)`

Through similar steps, you should find that another vertical asymptote is at
`x = -(\pi)/(6)`

We can condense those two equations into
`x = \pm(\pi)/(6)`