Find Where Undefined/Discontinuous tan(3x-2)^2

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Trigonometry

$\tan^{2} (3 x - 2)$

Set the argument in $\tan (3 x - 2)$ equal to $\frac{π}{2} + π n$ to find where the expression is undefined.

$3 x - 2 = \frac{π}{2} + π n$ , for any integer $n$

Solve for $x$ .

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Add $2$ to both sides of the equation.

$3 x = \frac{π}{2} + π n + 2$

Divide each term by $3$ and simplify.

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Divide each term in $3 x = \frac{π}{2} + π n + 2$ by $3$ .

$\frac{3 x}{3} = \frac{π}{2} \cdot \frac{1}{3} + \frac{π n}{3} + \frac{2}{3}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 x}{3} = \frac{π}{2} \cdot \frac{1}{3} + \frac{π n}{3} + \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{π}{2} \cdot \frac{1}{3} + \frac{π n}{3} + \frac{2}{3}$

Multiply $\frac{π}{2} \cdot \frac{1}{3}$ .

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Multiply $\frac{π}{2}$ and $\frac{1}{3}$ .

$x = \frac{π}{2 \cdot 3} + \frac{π n}{3} + \frac{2}{3}$

Multiply $2$ by $3$ .

$x = \frac{π}{6} + \frac{π n}{3} + \frac{2}{3}$

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

Set-Builder Notation:

${x | x = \frac{π}{6} + \frac{π n}{3} + \frac{2}{3}}$ , for any integer $n$

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