Calculus Examples

$y = \ln (\tan^{2} (x))$

Step 1

Set the argument in $\ln (\tan^{2} (x))$ less than or equal to $0$ to find where the expression is undefined.

$\tan^{2} (x) \leq 0$

Step 2

Solve for $x$ .

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Step 2.1

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{\tan^{2} (x)} \leq \sqrt{0}$

Step 2.2

Simplify the equation.

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Step 2.2.1

Simplify the left side.

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Step 2.2.1.1

Pull terms out from under the radical.

$| \tan (x) | \leq \sqrt{0}$

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

Simplify $\sqrt{0}$ .

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Step 2.2.2.1.1

Rewrite $0$ as $0^{2}$ .

$| \tan (x) | \leq \sqrt{0^{2}}$

Step 2.2.2.1.2

Pull terms out from under the radical.

$| \tan (x) | \leq | 0 |$

Step 2.2.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$| \tan (x) | \leq 0$

Step 2.3

Write $| \tan (x) | \leq 0$ as a piecewise.

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Step 2.3.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$\tan (x) \geq 0$

Step 2.3.2

In the piece where $\tan (x)$ is non-negative, remove the absolute value.

$\tan (x) \leq 0$

Step 2.3.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$\tan (x) < 0$

Step 2.3.4

In the piece where $\tan (x)$ is negative, remove the absolute value and multiply by $- 1$ .

$- \tan (x) \leq 0$

Step 2.3.5

Write as a piecewise.

${\begin{cases} \tan (x) \leq 0 & \tan (x) \geq 0 \\ - \tan (x) \leq 0 & \tan (x) < 0 \end{cases}$

Step 2.4

Find the intersection of $\tan (x) \leq 0$ and $\tan (x) \geq 0$ .

$\tan (x) = 0$

Step 2.5

Solve $- \tan (x) \leq 0$ when $\tan (x) < 0$ .

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Step 2.5.1

Divide each term in $- \tan (x) \leq 0$ by $- 1$ and simplify.

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Step 2.5.1.1

Divide each term in $- \tan (x) \leq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- \tan (x)}{- 1} \geq \frac{0}{- 1}$

Step 2.5.1.2

Simplify the left side.

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Step 2.5.1.2.1

Dividing two negative values results in a positive value.

$\frac{\tan (x)}{1} \geq \frac{0}{- 1}$

Step 2.5.1.2.2

Divide $\tan (x)$ by $1$ .

$\tan (x) \geq \frac{0}{- 1}$

Step 2.5.1.3

Simplify the right side.

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Step 2.5.1.3.1

Divide $0$ by $- 1$ .

$\tan (x) \geq 0$

Step 2.5.2

Find the intersection of $\tan (x) \geq 0$ and $\tan (x) < 0$ .

No solution

Step 2.6

Find the union of the solutions.

$\tan (x) = 0$

Step 2.7

Take the inverse tangent of both sides of the equation to extract $x$ from inside the tangent.

$x = \arctan (0)$

Step 2.8

Simplify the right side.

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Step 2.8.1

The exact value of $\arctan (0)$ is $0$ .

$x = 0$

Step 2.9

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = π + 0$

Step 2.10

Add $π$ and $0$ .

$x = π$

Step 2.11

Find the period of $\tan (x)$ .

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Step 2.11.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 2.11.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 2.11.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 2.11.4

Divide $π$ by $1$ .

$π$

Step 2.12

The period of the $\tan (x)$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = π n, π + π n$ , for any integer $n$

Step 2.13

Consolidate the answers.

$x = π n$ , for any integer $n$

Step 2.14

Find the domain of $\tan^{2} (x)$ .

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Step 2.14.1

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

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

Step 2.14.2

The domain is all values of $x$ that make the expression defined.

${x | x \neq \frac{π}{2} + π n}$ $n$ , for any integer $n$

Step 2.15

Use each root to create test intervals.

$0 < x < \frac{π}{2}$

$\frac{π}{2} < x < π$

Step 2.16

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.16.1

Test a value on the interval $0 < x < \frac{π}{2}$ to see if it makes the inequality true.

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Step 2.16.1.1

Choose a value on the interval $0 < x < \frac{π}{2}$ and see if this value makes the original inequality true.

$x = 1$

Step 2.16.1.2

Replace $x$ with $1$ in the original inequality.

$\tan^{2} (1) \leq 0$

Step 2.16.1.3

The left side $2.42551882$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 2.16.2

Test a value on the interval $\frac{π}{2} < x < π$ to see if it makes the inequality true.

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Step 2.16.2.1

Choose a value on the interval $\frac{π}{2} < x < π$ and see if this value makes the original inequality true.

$x = 2$

Step 2.16.2.2

Replace $x$ with $2$ in the original inequality.

$\tan^{2} (2) \leq 0$

Step 2.16.2.3

The left side $4.7743992$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 2.16.3

Compare the intervals to determine which ones satisfy the original inequality.

$0 < x < \frac{π}{2}$ False

$\frac{π}{2} < x < π$ False

$0 < x < \frac{π}{2}$ False

$\frac{π}{2} < x < π$ False

Step 2.17

Since there are no numbers that fall within the interval, this inequality has no solution.

No solution

Step 3

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

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

Step 4

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 $0$ .

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

Step 5