Finite Math Examples

$\frac{\sin (x)}{\sin (x) + 1} > 1$

Step 1

Subtract $1$ from both sides of the inequality.

$\frac{\sin (x)}{\sin (x) + 1} - 1 > 0$

Step 2

Simplify $\frac{\sin (x)}{\sin (x) + 1} - 1$ .

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Step 2.1

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{\sin (x) + 1}{\sin (x) + 1}$ .

$\frac{\sin (x)}{\sin (x) + 1} - 1 \cdot \frac{\sin (x) + 1}{\sin (x) + 1} > 0$

Step 2.2

Combine $- 1$ and $\frac{\sin (x) + 1}{\sin (x) + 1}$ .

$\frac{\sin (x)}{\sin (x) + 1} + \frac{- (\sin (x) + 1)}{\sin (x) + 1} > 0$

Step 2.3

Combine the numerators over the common denominator.

$\frac{\sin (x) - (\sin (x) + 1)}{\sin (x) + 1} > 0$

Step 2.4

Simplify the numerator.

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Step 2.4.1

Apply the distributive property.

$\frac{\sin (x) - \sin (x) - 1 \cdot 1}{\sin (x) + 1} > 0$

Step 2.4.2

Multiply $- 1$ by $1$ .

$\frac{\sin (x) - \sin (x) - 1}{\sin (x) + 1} > 0$

Step 2.4.3

Subtract $\sin (x)$ from $\sin (x)$ .

$\frac{0 - 1}{\sin (x) + 1} > 0$

Step 2.4.4

Subtract $1$ from $0$ .

$\frac{- 1}{\sin (x) + 1} > 0$

Step 2.5

Move the negative in front of the fraction.

$- \frac{1}{\sin (x) + 1} > 0$

Step 3

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$\sin (x) + 1 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

Step 5

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

$x = \arcsin (- 1)$

Step 6

Simplify the right side.

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Step 6.1

The exact value of $\arcsin (- 1)$ is $- \frac{π}{2}$ .

$x = - \frac{π}{2}$

Step 7

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{2} + π$

Step 8

Simplify the expression to find the second solution.

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Step 8.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 8.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 9

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

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Step 9.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 9.2

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

$\frac{2 π}{| 1 |}$

Step 9.3

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

$\frac{2 π}{1}$

Step 9.4

Divide $2 π$ by $1$ .

$2 π$

Step 10

Add $2 π$ to every negative angle to get positive angles.

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Step 10.1

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

Step 10.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 10.3

Combine fractions.

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Step 10.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 10.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 10.4

Simplify the numerator.

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Step 10.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 10.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 10.5

List the new angles.

$x = \frac{3 π}{2}$

Step 11

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

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

Step 12

Consolidate the answers.

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

Step 13

Find the domain of $\frac{\sin (x)}{\sin (x) + 1} - 1$ .

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Step 13.1

Set the denominator in $\frac{\sin (x)}{\sin (x) + 1}$ equal to $0$ to find where the expression is undefined.

$\sin (x) + 1 = 0$

Step 13.2

Solve for $x$ .

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Step 13.2.1

Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

Step 13.2.2

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

$x = \arcsin (- 1)$

Step 13.2.3

Simplify the right side.

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Step 13.2.3.1

The exact value of $\arcsin (- 1)$ is $- \frac{π}{2}$ .

$x = - \frac{π}{2}$

Step 13.2.4

$x = 2 π + \frac{π}{2} + π$

Step 13.2.5

Simplify the expression to find the second solution.

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Step 13.2.5.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 13.2.5.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 13.2.6

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

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Step 13.2.6.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 13.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 13.2.6.3

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

$\frac{2 π}{1}$

Step 13.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 13.2.7

Add $2 π$ to every negative angle to get positive angles.

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Step 13.2.7.1

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

Step 13.2.7.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 13.2.7.3

Combine fractions.

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Step 13.2.7.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 13.2.7.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 13.2.7.4

Simplify the numerator.

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Step 13.2.7.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 13.2.7.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 13.2.7.5

List the new angles.

$x = \frac{3 π}{2}$

Step 13.2.8

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

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

Step 13.2.9

Consolidate the answers.

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

Step 13.3

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

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

Step 14

Use each root to create test intervals.

$\frac{3 π}{2} < x < \frac{7 π}{2}$

Step 15

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 15.1

Test a value on the interval $\frac{3 π}{2} < x < \frac{7 π}{2}$ to see if it makes the inequality true.

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Step 15.1.1

Choose a value on the interval $\frac{3 π}{2} < x < \frac{7 π}{2}$ and see if this value makes the original inequality true.

$x = 8$

Step 15.1.2

Replace $x$ with $8$ in the original inequality.

$\frac{\sin (8)}{\sin (8) + 1} > 1$

Step 15.1.3

The left side $0.49732533$ is not greater than the right side $1$ , which means that the given statement is false.

False

Step 15.2

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

$\frac{3 π}{2} < x < \frac{7 π}{2}$ False

Step 16

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

No solution