Precalculus Examples

$\tan (x) < 0$ , $\csc (x) < 0$

Step 1

Simplify the first inequality.

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Step 1.1

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

$x < \arctan (0)$ and $\csc (x) < 0$

Step 1.2

Simplify the right side.

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Step 1.2.1

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

$x < 0$ and $\csc (x) < 0$

Step 1.3

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.

$x = π + 0$ and $\csc (x) < 0$

Step 1.4

Add $π$ and $0$ .

$x = π$ and $\csc (x) < 0$

Step 1.5

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

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Step 1.5.1

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

$\frac{π}{| b |}$

Step 1.5.2

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

$\frac{π}{| 1 |}$

Step 1.5.3

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

$\frac{π}{1}$

Step 1.5.4

Divide $π$ by $1$ .

$π$

Step 1.6

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

$x = π n, π + π n$ and $\csc (x) < 0$

Step 1.7

Consolidate the answers.

$x = π n$ and $\csc (x) < 0$

Step 1.8

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

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Step 1.8.1

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 1.8.2

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 1.9

Use each root to create test intervals.

$0 < x < \frac{π}{2}$

$\frac{π}{2} < x < π$ and $\csc (x) < 0$

Step 1.10

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 1.10.1

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

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Step 1.10.1.1

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

$x = 1$ and $\csc (x) < 0$

Step 1.10.1.2

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

$\tan (1) < 0$ and $\csc (x) < 0$

Step 1.10.1.3

The left side $1.55740772$ is not less than the right side $0$ , which means that the given statement is false.

False and $\csc (x) < 0$

Step 1.10.2

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

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Step 1.10.2.1

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

$x = 2$ and $\csc (x) < 0$

Step 1.10.2.2

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

$\tan (2) < 0$ and $\csc (x) < 0$

Step 1.10.2.3

The left side $- 2.18503986$ is less than the right side $0$ , which means that the given statement is always true.

True and $\csc (x) < 0$

Step 1.10.3

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

$0 < x < \frac{π}{2}$ False

$\frac{π}{2} < x < π$ True and $\csc (x) < 0$

$0 < x < \frac{π}{2}$ False

$\frac{π}{2} < x < π$ True and $\csc (x) < 0$

Step 1.11

The solution consists of all of the true intervals.

$\frac{π}{2} + π n < x < π + π n$ and $\csc (x) < 0$

Step 2

The range of cosecant is $y \leq - 1$ and $y \geq 1$ . Since $0$ does not fall in this range, there is no solution.

No solution