Algebra Examples

$f (x) = \arcsin (\cos (x))$

Step 1

Set the argument in $\arcsin (\cos (x))$ greater than or equal to $- 1$ to find where the expression is defined.

$\cos (x) \geq - 1$

Step 2

Solve for $x$ .

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Step 2.1

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

$x \geq \arccos (- 1)$

Step 2.2

Simplify the right side.

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Step 2.2.1

The exact value of $\arccos (- 1)$ is $π$ .

$x \geq π$

Step 2.3

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 π - π$

Step 2.4

Subtract $π$ from $2 π$ .

$x = π$

Step 2.5

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

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Step 2.5.1

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

$\frac{2 π}{| b |}$

Step 2.5.2

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

$\frac{2 π}{| 1 |}$

Step 2.5.3

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

$\frac{2 π}{1}$

Step 2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.6

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

$x = π + 2 π n$ , for any integer $n$

Step 2.7

Use each root to create test intervals.

$π < x < 3 π$

Step 2.8

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 2.8.1

Test a value on the interval $π < x < 3 π$ to see if it makes the inequality true.

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Step 2.8.1.1

Choose a value on the interval $π < x < 3 π$ and see if this value makes the original inequality true.

$x = 6$

Step 2.8.1.2

Replace $x$ with $6$ in the original inequality.

$\cos (6) \geq - 1$

Step 2.8.1.3

The left side $0.96017028$ is greater than the right side $- 1$ , which means that the given statement is always true.

True

Step 2.8.2

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

$π < x < 3 π$ True

Step 2.9

The solution consists of all of the true intervals.

$π + 2 π n \leq x \leq 3 π + 2 π n$ , for any integer $n$

Step 3

Set the argument in $\arcsin (\cos (x))$ less than or equal to $1$ to find where the expression is defined.

$\cos (x) \leq 1$

Step 4

Solve for $x$ .

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Step 4.1

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

$x \leq \arccos (1)$

Step 4.2

Simplify the right side.

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Step 4.2.1

The exact value of $\arccos (1)$ is $0$ .

$x \leq 0$

Step 4.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - 0$

Step 4.4

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 4.5

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

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Step 4.5.1

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

$\frac{2 π}{| b |}$

Step 4.5.2

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

$\frac{2 π}{| 1 |}$

Step 4.5.3

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

$\frac{2 π}{1}$

Step 4.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.6

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

$x = 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 4.7

Consolidate the answers.

$x = 2 π n$ , for any integer $n$

Step 4.8

Use each root to create test intervals.

$0 < x < 2 π$

Step 4.9

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 4.9.1

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

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Step 4.9.1.1

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

$x = 3$

Step 4.9.1.2

Replace $x$ with $3$ in the original inequality.

$\cos (3) \leq 1$

Step 4.9.1.3

The left side $- 0.98999249$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 4.9.2

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

$0 < x < 2 π$ True

Step 4.10

The solution consists of all of the true intervals.

$0 + 2 π n \leq x \leq 2 π + 2 π n$ , for any integer $n$

Step 5

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

Set-Builder Notation:

${x | x = π + 2 π n, x = 3 π + 2 π n, x = 0 + 2 π n, x = 2 π + 2 π n}$ , for any integer $n$

Step 6

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[- \frac{π}{2}, \frac{π}{2}]$

Set-Builder Notation:

${y | - \frac{π}{2} \leq y \leq \frac{π}{2}}$

Step 7

Determine the domain and range.

Domain: ${x | x = π + 2 π n, x = 3 π + 2 π n, x = 0 + 2 π n, x = 2 π + 2 π n}$ , for any integer $n$

Range: $[- \frac{π}{2}, \frac{π}{2}], {y | - \frac{π}{2} \leq y \leq \frac{π}{2}}$

Step 8