Precalculus Examples

cos (2 x) - cos (x) = 0

$\cos (2 x) - \cos (x) = 0$

Step 1

Use the double-angle identity to transform

cos (2 x)

$\cos (2 x)$ to

2 {cos}^{2} (x) - 1

$2 \cos^{2} (x) - 1$ .

2 {cos}^{2} (x) - 1 - cos (x) = 0

$2 \cos^{2} (x) - 1 - \cos (x) = 0$

Step 2

Factor by grouping.

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Step 2.1

Reorder terms.

2 {cos}^{2} (x) - cos (x) - 1 = 0

$2 \cos^{2} (x) - \cos (x) - 1 = 0$

Step 2.2

For a polynomial of the form

a x^{2} + b x + c

$a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is

a \cdot c = 2 \cdot - 1 = - 2

$a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is

b = - 1

$b = - 1$ .

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Step 2.2.1

Factor

- 1

$- 1$ out of

- cos (x)

$- \cos (x)$ .

2 {cos}^{2} (x) - cos (x) - 1 = 0

$2 \cos^{2} (x) - \cos (x) - 1 = 0$

Step 2.2.2

Rewrite

- 1

$- 1$ as

1

$1$ plus

- 2

$- 2$

2 {cos}^{2} (x) + (1 - 2) cos (x) - 1 = 0

$2 \cos^{2} (x) + (1 - 2) \cos (x) - 1 = 0$

Step 2.2.3

Apply the distributive property.

2 {cos}^{2} (x) + 1 cos (x) - 2 cos (x) - 1 = 0

$2 \cos^{2} (x) + 1 \cos (x) - 2 \cos (x) - 1 = 0$

Step 2.2.4

Multiply

cos (x)

$\cos (x)$ by

1

$1$ .

2 {cos}^{2} (x) + cos (x) - 2 cos (x) - 1 = 0

$2 \cos^{2} (x) + \cos (x) - 2 \cos (x) - 1 = 0$

2 {cos}^{2} (x) + cos (x) - 2 cos (x) - 1 = 0

$2 \cos^{2} (x) + \cos (x) - 2 \cos (x) - 1 = 0$

Step 2.3

Factor out the greatest common factor from each group.

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Step 2.3.1

Group the first two terms and the last two terms.

2 {cos}^{2} (x) + cos (x) - 2 cos (x) - 1 = 0

$2 \cos^{2} (x) + \cos (x) - 2 \cos (x) - 1 = 0$

Step 2.3.2

Factor out the greatest common factor (GCF) from each group.

cos (x) (2 cos (x) + 1) - (2 cos (x) + 1) = 0

$\cos (x) (2 \cos (x) + 1) - (2 \cos (x) + 1) = 0$

cos (x) (2 cos (x) + 1) - (2 cos (x) + 1) = 0

$\cos (x) (2 \cos (x) + 1) - (2 \cos (x) + 1) = 0$

Step 2.4

Factor the polynomial by factoring out the greatest common factor,

2 cos (x) + 1

$2 \cos (x) + 1$ .

(2 cos (x) + 1) (cos (x) - 1) = 0

$(2 \cos (x) + 1) (\cos (x) - 1) = 0$

(2 cos (x) + 1) (cos (x) - 1) = 0

$(2 \cos (x) + 1) (\cos (x) - 1) = 0$

Step 3

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

2 cos (x) + 1 = 0

$2 \cos (x) + 1 = 0$

cos (x) - 1 = 0

$\cos (x) - 1 = 0$

Step 4

Set

2 cos (x) + 1

$2 \cos (x) + 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 4.1

Set

2 cos (x) + 1

$2 \cos (x) + 1$ equal to

0

$0$ .

2 cos (x) + 1 = 0

$2 \cos (x) + 1 = 0$

Step 4.2

Solve

2 cos (x) + 1 = 0

$2 \cos (x) + 1 = 0$ for

x

$x$ .

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Step 4.2.1

Subtract

1

$1$ from both sides of the equation.

2 cos (x) = - 1

$2 \cos (x) = - 1$

Step 4.2.2

Divide each term in

2 cos (x) = - 1

$2 \cos (x) = - 1$ by

2

$2$ and simplify.

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Step 4.2.2.1

Divide each term in

2 cos (x) = - 1

$2 \cos (x) = - 1$ by

2

$2$ .

\frac{2 cos (x)}{2} = \frac{- 1}{2}

$\frac{2 \cos (x)}{2} = \frac{- 1}{2}$

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of

2

$2$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

\frac{2 cos (x)}{2} = \frac{- 1}{2}

$\frac{2 \cos (x)}{2} = \frac{- 1}{2}$

Step 4.2.2.2.1.2

Divide

cos (x)

$\cos (x)$ by

1

$1$ .

cos (x) = \frac{- 1}{2}

$\cos (x) = \frac{- 1}{2}$

cos (x) = \frac{- 1}{2}

$\cos (x) = \frac{- 1}{2}$

cos (x) = \frac{- 1}{2}

$\cos (x) = \frac{- 1}{2}$

Step 4.2.2.3

Simplify the right side.

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Step 4.2.2.3.1

Move the negative in front of the fraction.

cos (x) = - \frac{1}{2}

$\cos (x) = - \frac{1}{2}$

cos (x) = - \frac{1}{2}

$\cos (x) = - \frac{1}{2}$

cos (x) = - \frac{1}{2}

$\cos (x) = - \frac{1}{2}$

Step 4.2.3

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

x

$x$ from inside the cosine.

x = arccos (- \frac{1}{2})

$x = \arccos (- \frac{1}{2})$

Step 4.2.4

Simplify the right side.

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Step 4.2.4.1

The exact value of

arccos (- \frac{1}{2})

$\arccos (- \frac{1}{2})$ is

\frac{2 π}{3}

$\frac{2 π}{3}$ .

x = \frac{2 π}{3}

$x = \frac{2 π}{3}$

x = \frac{2 π}{3}

$x = \frac{2 π}{3}$

Step 4.2.5

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 - \frac{2 π}{3}

$x = 2 π - \frac{2 π}{3}$

Step 4.2.6

Simplify

2 π - \frac{2 π}{3}

$2 π - \frac{2 π}{3}$ .

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Step 4.2.6.1

To write

2 π

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

\frac{3}{3}

$\frac{3}{3}$ .

x = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}

$x = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Step 4.2.6.2

Combine fractions.

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Step 4.2.6.2.1

Combine

2 π

$2 π$ and

\frac{3}{3}

$\frac{3}{3}$ .

x = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}

$x = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}$

Step 4.2.6.2.2

Combine the numerators over the common denominator.

x = \frac{2 π \cdot 3 - 2 π}{3}

$x = \frac{2 π \cdot 3 - 2 π}{3}$

x = \frac{2 π \cdot 3 - 2 π}{3}

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 4.2.6.3

Simplify the numerator.

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Step 4.2.6.3.1

Multiply

3

$3$ by

2

$2$ .

x = \frac{6 π - 2 π}{3}

$x = \frac{6 π - 2 π}{3}$

Step 4.2.6.3.2

Subtract

2 π

$2 π$ from

6 π

$6 π$ .

x = \frac{4 π}{3}

$x = \frac{4 π}{3}$

x = \frac{4 π}{3}

$x = \frac{4 π}{3}$

x = \frac{4 π}{3}

$x = \frac{4 π}{3}$

Step 4.2.7

Find the period of

cos (x)

$\cos (x)$ .

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Step 4.2.7.1

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 4.2.7.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 4.2.7.3

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 4.2.7.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 4.2.8

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{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer

n

$n$

x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer

n

$n$

x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer

n

$n$

Step 5

Set

cos (x) - 1

$\cos (x) - 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 5.1

Set

cos (x) - 1

$\cos (x) - 1$ equal to

0

$0$ .

cos (x) - 1 = 0

$\cos (x) - 1 = 0$

Step 5.2

Solve

cos (x) - 1 = 0

$\cos (x) - 1 = 0$ for

x

$x$ .

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Step 5.2.1

Add

1

$1$ to both sides of the equation.

cos (x) = 1

$\cos (x) = 1$

Step 5.2.2

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

x

$x$ from inside the cosine.

x = arccos (1)

$x = \arccos (1)$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

The exact value of

arccos (1)

$\arccos (1)$ is

0

$0$ .

x = 0

$x = 0$

x = 0

$x = 0$

Step 5.2.4

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 - 0

$x = 2 π - 0$

Step 5.2.5

Subtract

0

$0$ from

2 π

$2 π$ .

x = 2 π

$x = 2 π$

Step 5.2.6

Find the period of

cos (x)

$\cos (x)$ .

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Step 5.2.6.1

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 5.2.6.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 5.2.6.3

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 5.2.6.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 5.2.7

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 = 2 π n, 2 π + 2 π n

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

n

$n$

x = 2 π n, 2 π + 2 π n

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

n

$n$

x = 2 π n, 2 π + 2 π n

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

n

$n$

Step 6

The final solution is all the values that make

(2 cos (x) + 1) (cos (x) - 1) = 0

$(2 \cos (x) + 1) (\cos (x) - 1) = 0$ true.

x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, 2 π n, 2 π + 2 π n

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, 2 π n, 2 π + 2 π n$ , for any integer

n

$n$

Step 7

Consolidate the answers.

x = \frac{2 π n}{3}

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

n

$n$

Step 8