Question

High school math question

Answer 1

Answer: Choice C. $x = \pm\frac{\pi}{6}$

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Explanation:

Recall that

$\tan(x) = \frac{\sin(x)}{\cos(x)}$

So we can say

$\tan(3x) = \frac{\sin(3x)}{\cos(3x)}$

The vertical asymptotes occur when the denominator is zero.

If we were to plug in $x = \frac{\pi}{6}$, then we'd have,

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

This shows that one vertical asymptote is at $x = \frac{\pi}{6}$

Through similar steps, you should find that another vertical asymptote is at $x = -\frac{\pi}{6}$

We can condense those two equations into $x = \pm\frac{\pi}{6}$