Algebra Examples

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

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

Step 1

Set the argument in

\arcsin (\cos (x))

$\arcsin (\cos (x))$ greater than or equal to

- 1

$- 1$ to find where the expression is defined.

\cos (x) \geq - 1

$\cos (x) \geq - 1$

Step 2

Solve for

x

$x$ .

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Step 2.1

Take the inverse cosine of both sides of the equation to extract

x

$x$ from inside the cosine.

x \geq \arccos (- 1)

$x \geq \arccos (- 1)$

Step 2.2

Simplify the right side.

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Step 2.2.1

The exact value of

\arccos (- 1)

$\arccos (- 1)$ is

π

$π$ .

x \geq π

$x \geq π$

x \geq 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

$2 π$ to find the solution in the third quadrant.

x = 2 π - π

$x = 2 π - π$

Step 2.4

Subtract

π

$π$ from

2 π

$2 π$ .

x = π

$x = π$

Step 2.5

Find the period of

\cos (x)

$\cos (x)$ .

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Step 2.5.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 2.5.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 2.5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 2.6

The period of the

\cos (x)

$\cos (x)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = π + 2 π n

$x = π + 2 π n$ , for any integer

n

$n$

Step 2.7

Use each root to create test intervals.

π < x < 3 π

$π < x < 3 π$

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

$π < x < 3 π$ 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

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

x = 6

$x = 6$

Step 2.8.1.2

Replace

x

$x$ with

6

$6$ in the original inequality.

\cos (6) \geq - 1

$\cos (6) \geq - 1$

Step 2.8.1.3

The left side

0.96017028

$0.96017028$ is greater than the right side

- 1

$- 1$ , 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

$π < x < 3 π$ True

π < x < 3 π

$π < x < 3 π$ 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

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

n

$n$

π + 2 π n \leq x \leq 3 π + 2 π n

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

n

$n$

Step 3

Set the argument in

\arcsin (\cos (x))

$\arcsin (\cos (x))$ less than or equal to

1

$1$ to find where the expression is defined.

\cos (x) \leq 1

$\cos (x) \leq 1$

Step 4

Solve for

x

$x$ .

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Step 4.1

Take the inverse cosine of both sides of the equation to extract

x

$x$ from inside the cosine.

x \leq \arccos (1)

$x \leq \arccos (1)$

Step 4.2

Simplify the right side.

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Step 4.2.1

The exact value of

\arccos (1)

$\arccos (1)$ is

0

$0$ .

x \leq 0

$x \leq 0$

x \leq 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

$2 π$ to find the solution in the fourth quadrant.

x = 2 π - 0

$x = 2 π - 0$

Step 4.4

Subtract

0

$0$ from

2 π

$2 π$ .

x = 2 π

$x = 2 π$

Step 4.5

Find the period of

\cos (x)

$\cos (x)$ .

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Step 4.5.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 4.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 4.5.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 4.5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 4.6

The period of the

\cos (x)

$\cos (x)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = 2 π n, 2 π + 2 π n

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

n

$n$

Step 4.7

Consolidate the answers.

x = 2 π n

$x = 2 π n$ , for any integer

n

$n$

Step 4.8

Use each root to create test intervals.

0 < x < 2 π

$0 < x < 2 π$

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

$0 < x < 2 π$ 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

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

x = 3

$x = 3$

Step 4.9.1.2

Replace

x

$x$ with

3

$3$ in the original inequality.

\cos (3) \leq 1

$\cos (3) \leq 1$

Step 4.9.1.3

The left side

- 0.98999249

$- 0.98999249$ is less than the right side

1

$1$ , 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

$0 < x < 2 π$ True

0 < x < 2 π

$0 < x < 2 π$ 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

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

n

$n$

0 + 2 π n \leq x \leq 2 π + 2 π n

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

n

$n$

Step 5

The domain is all values of

x

$x$ 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}

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

n

$n$

Step 6

The range is the set of all valid

y

$y$ values. Use the graph to find the range.

Interval Notation:

[- \frac{π}{2}, \frac{π}{2}]

$[- \frac{π}{2}, \frac{π}{2}]$

Set-Builder Notation:

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

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

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}

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

n

$n$

Range:

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

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

Step 8