Precalculus Examples

y = cos (x)

$y = \cos (x)$

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in

0

$0$ for

y

$y$ and solve for

x

$x$ .

0 = cos (x)

$0 = \cos (x)$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as

cos (x) = 0

$\cos (x) = 0$ .

cos (x) = 0

$\cos (x) = 0$

Step 1.2.2

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

x

$x$ from inside the cosine.

x = arccos (0)

$x = \arccos (0)$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

The exact value of

arccos (0)

$\arccos (0)$ is

\frac{π}{2}

$\frac{π}{2}$ .

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

Step 1.2.4

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 - \frac{π}{2}

$x = 2 π - \frac{π}{2}$

Step 1.2.5

Simplify

2 π - \frac{π}{2}

$2 π - \frac{π}{2}$ .

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Step 1.2.5.1

To write

2 π

$2 π$ as a fraction with a common denominator, multiply by

\frac{2}{2}

$\frac{2}{2}$ .

x = 2 π \cdot \frac{2}{2} - \frac{π}{2}

$x = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 1.2.5.2

Combine fractions.

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Step 1.2.5.2.1

Combine

2 π

$2 π$ and

\frac{2}{2}

$\frac{2}{2}$ .

x = \frac{2 π \cdot 2}{2} - \frac{π}{2}

$x = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 1.2.5.2.2

Combine the numerators over the common denominator.

x = \frac{2 π \cdot 2 - π}{2}

$x = \frac{2 π \cdot 2 - π}{2}$

x = \frac{2 π \cdot 2 - π}{2}

$x = \frac{2 π \cdot 2 - π}{2}$

Step 1.2.5.3

Simplify the numerator.

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Step 1.2.5.3.1

Multiply

2

$2$ by

2

$2$ .

x = \frac{4 π - π}{2}

$x = \frac{4 π - π}{2}$

Step 1.2.5.3.2

Subtract

π

$π$ from

4 π

$4 π$ .

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

Step 1.2.6

Find the period of

cos (x)

$\cos (x)$ .

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Step 1.2.6.1

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 1.2.6.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 1.2.6.3

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 1.2.6.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 1.2.7

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 = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n

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

$n$

Step 1.2.8

Consolidate the answers.

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

$n$

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

$n$

Step 1.3

x-intercept(s) in point form.

x-intercept(s):

$(\frac{π}{2} + π n, 0)$ , for any integer

$n$

x-intercept(s):

$(\frac{π}{2} + π n, 0)$ , for any integer

$n$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in

$0$ for

$x$ and solve for

$y$ .

$y = \cos (0)$

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

$y = \cos (0)$

Step 2.2.2

The exact value of

$\cos (0)$ is

$1$ .

$y = 1$

Step 2.3

y-intercept(s) in point form.

y-intercept(s):

$(0, 1)$

y-intercept(s):

$(0, 1)$

Step 3

List the intersections.

x-intercept(s):

$(\frac{π}{2} + π n, 0)$ , for any integer

$n$

y-intercept(s):

$(0, 1)$

Step 4