Algebra Examples

$\sin (x) > - 1$

Step 1

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

$x > \arcsin (- 1)$

Step 2

Simplify the right side.

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Step 2.1

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

$x > - \frac{π}{2}$

Step 3

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 4

Simplify the expression to find the second solution.

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Step 4.1

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

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 4.2

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 5

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

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Step 5.1

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

$\frac{2 π}{| b |}$

Step 5.2

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

$\frac{2 π}{| 1 |}$

Step 5.3

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

$\frac{2 π}{1}$

Step 5.4

Divide $2 π$ by $1$ .

$2 π$

Step 6

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

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Step 6.1

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

$- \frac{π}{2} + 2 π$

Step 6.2

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 6.3

Combine fractions.

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Step 6.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 6.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 6.4

Simplify the numerator.

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Step 6.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 6.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 6.5

List the new angles.

$x = \frac{3 π}{2}$

Step 7

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 8

Consolidate the answers.

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

Step 9

Use each root to create test intervals.

$\frac{3 π}{2} < x < \frac{7 π}{2}$

Step 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 10.1

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 10.1.1

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 10.1.2

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

$\sin (8) > - 1$

Step 10.1.3

The left side $0.98935824$ is greater than the right side $- 1$ , which means that the given statement is always true.

True

Step 10.2

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

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

Step 11

The solution consists of all of the true intervals.

$\frac{3 π}{2} + 2 π n < x < \frac{7 π}{2} + 2 π n$ , for any integer $n$

Step 12