Precalculus Examples

$\cos (4 x) - \cos (2 x) = 0$

Step 1

Simplify the left side of the equation.

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Step 1.1

Simplify each term.

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Step 1.1.1

Factor $2$ out of $4 x$ .

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

Step 1.1.2

Simplify each term.

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Step 1.1.2.1

Use the double-angle identity to transform $\cos (2 x)$ to $2 \cos^{2} (x) - 1$ .

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Expand $(2 \cos^{2} (x) - 1) (2 \cos^{2} (x) - 1)$ using the FOIL Method.

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Step 1.1.2.3.1

Apply the distributive property.

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

Step 1.1.2.3.2

Apply the distributive property.

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

Step 1.1.2.3.3

Apply the distributive property.

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

Step 1.1.2.4

Simplify and combine like terms.

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Step 1.1.2.4.1

Simplify each term.

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Step 1.1.2.4.1.1

Rewrite using the commutative property of multiplication.

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

Step 1.1.2.4.1.2

Multiply $\cos^{2} (x)$ by $\cos^{2} (x)$ by adding the exponents.

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Step 1.1.2.4.1.2.1

Move $\cos^{2} (x)$ .

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

Step 1.1.2.4.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 1.1.2.4.1.2.3

Add $2$ and $2$ .

$2 (2 \cdot (2 \cos^{4} (x)) + 2 \cos^{2} (x) \cdot - 1 - 1 (2 \cos^{2} (x)) - 1 \cdot - 1) - 1 - \cos (2 x) = 0$

Step 1.1.2.4.1.3

Multiply $2$ by $2$ .

$2 (4 \cos^{4} (x) + 2 \cos^{2} (x) \cdot - 1 - 1 (2 \cos^{2} (x)) - 1 \cdot - 1) - 1 - \cos (2 x) = 0$

Step 1.1.2.4.1.4

Multiply $- 1$ by $2$ .

$2 (4 \cos^{4} (x) - 2 \cos^{2} (x) - 1 (2 \cos^{2} (x)) - 1 \cdot - 1) - 1 - \cos (2 x) = 0$

Step 1.1.2.4.1.5

Multiply $2$ by $- 1$ .

$2 (4 \cos^{4} (x) - 2 \cos^{2} (x) - 2 \cos^{2} (x) - 1 \cdot - 1) - 1 - \cos (2 x) = 0$

Step 1.1.2.4.1.6

Multiply $- 1$ by $- 1$ .

$2 (4 \cos^{4} (x) - 2 \cos^{2} (x) - 2 \cos^{2} (x) + 1) - 1 - \cos (2 x) = 0$

Step 1.1.2.4.2

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

$2 (4 \cos^{4} (x) - 4 \cos^{2} (x) + 1) - 1 - \cos (2 x) = 0$

Step 1.1.2.5

Apply the distributive property.

$2 (4 \cos^{4} (x)) + 2 (- 4 \cos^{2} (x)) + 2 \cdot 1 - 1 - \cos (2 x) = 0$

Step 1.1.2.6

Simplify.

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Step 1.1.2.6.1

Multiply $4$ by $2$ .

$8 \cos^{4} (x) + 2 (- 4 \cos^{2} (x)) + 2 \cdot 1 - 1 - \cos (2 x) = 0$

Step 1.1.2.6.2

Multiply $- 4$ by $2$ .

$8 \cos^{4} (x) - 8 \cos^{2} (x) + 2 \cdot 1 - 1 - \cos (2 x) = 0$

Step 1.1.2.6.3

Multiply $2$ by $1$ .

$8 \cos^{4} (x) - 8 \cos^{2} (x) + 2 - 1 - \cos (2 x) = 0$

Step 1.1.3

Subtract $1$ from $2$ .

$8 \cos^{4} (x) - 8 \cos^{2} (x) + 1 - \cos (2 x) = 0$

Step 1.1.4

Use the double-angle identity to transform $\cos (2 x)$ to $2 \cos^{2} (x) - 1$ .

$8 \cos^{4} (x) - 8 \cos^{2} (x) + 1 - (2 \cos^{2} (x) - 1) = 0$

Step 1.1.5

Apply the distributive property.

$8 \cos^{4} (x) - 8 \cos^{2} (x) + 1 - (2 \cos^{2} (x)) + 1 = 0$

Step 1.1.6

Multiply $2$ by $- 1$ .

$8 \cos^{4} (x) - 8 \cos^{2} (x) + 1 - 2 \cos^{2} (x) + 1 = 0$

Step 1.1.7

Multiply $- 1$ by $- 1$ .

$8 \cos^{4} (x) - 8 \cos^{2} (x) + 1 - 2 \cos^{2} (x) + 1 = 0$

Step 1.2

Simplify by adding terms.

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Step 1.2.1

Subtract $2 \cos^{2} (x)$ from $- 8 \cos^{2} (x)$ .

$8 \cos^{4} (x) + 1 - 10 \cos^{2} (x) + 1 = 0$

Step 1.2.2

Add $1$ and $1$ .

$8 \cos^{4} (x) + 2 - 10 \cos^{2} (x) = 0$

Step 2

Factor $8 \cos^{4} (x) + 2 - 10 \cos^{2} (x)$ .

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Step 2.1

Factor $2$ out of $8 \cos^{4} (x) + 2 - 10 \cos^{2} (x)$ .

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Step 2.1.1

Factor $2$ out of $8 \cos^{4} (x)$ .

$2 (4 \cos^{4} (x)) + 2 - 10 \cos^{2} (x) = 0$

Step 2.1.2

Factor $2$ out of $2$ .

$2 (4 \cos^{4} (x)) + 2 (1) - 10 \cos^{2} (x) = 0$

Step 2.1.3

Factor $2$ out of $- 10 \cos^{2} (x)$ .

$2 (4 \cos^{4} (x)) + 2 (1) + 2 (- 5 \cos^{2} (x)) = 0$

Step 2.1.4

Factor $2$ out of $2 (4 \cos^{4} (x)) + 2 (1)$ .

$2 (4 \cos^{4} (x) + 1) + 2 (- 5 \cos^{2} (x)) = 0$

Step 2.1.5

Factor $2$ out of $2 (4 \cos^{4} (x) + 1) + 2 (- 5 \cos^{2} (x))$ .

$2 (4 \cos^{4} (x) + 1 - 5 \cos^{2} (x)) = 0$

Step 2.2

Factor by grouping.

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Step 2.2.1

Reorder terms.

$2 (4 \cos^{4} (x) - 5 \cos^{2} (x) + 1) = 0$

Step 2.2.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot 1 = 4$ and whose sum is $b = - 5$ .

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Step 2.2.2.1

Factor $- 5$ out of $- 5 \cos^{2} (x)$ .

$2 (4 \cos^{4} (x) - 5 \cos^{2} (x) + 1) = 0$

Step 2.2.2.2

Rewrite $- 5$ as $- 1$ plus $- 4$

$2 (4 \cos^{4} (x) + (- 1 - 4) \cos^{2} (x) + 1) = 0$

Step 2.2.2.3

Apply the distributive property.

$2 (4 \cos^{4} (x) - 1 \cos^{2} (x) - 4 \cos^{2} (x) + 1) = 0$

Step 2.2.3

Factor out the greatest common factor from each group.

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Step 2.2.3.1

Group the first two terms and the last two terms.

$2 (4 \cos^{4} (x) - 1 \cos^{2} (x) - 4 \cos^{2} (x) + 1) = 0$

Step 2.2.3.2

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

$2 (\cos^{2} (x) (4 \cos^{2} (x) - 1) - (4 \cos^{2} (x) - 1)) = 0$

Step 2.2.4

Factor the polynomial by factoring out the greatest common factor, $4 \cos^{2} (x) - 1$ .

$2 ((4 \cos^{2} (x) - 1) (\cos^{2} (x) - 1)) = 0$

Step 2.3

Rewrite $4 \cos^{2} (x)$ as ${(2 \cos (x))}^{2}$ .

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

Step 2.4

Rewrite $1$ as $1^{2}$ .

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

Step 2.5

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 \cos (x)$ and $b = 1$ .

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

Step 2.6

Rewrite $1$ as $1^{2}$ .

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

Step 2.7

Factor.

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Step 2.7.1

Factor.

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Step 2.7.1.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \cos (x)$ and $b = 1$ .

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

Step 2.7.1.2

Remove unnecessary parentheses.

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

Step 2.7.2

Remove unnecessary parentheses.

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

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

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

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

$\cos (x) + 1 = 0$

$\cos (x) - 1 = 0$

Step 4

Set $2 \cos (x) + 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $2 \cos (x) + 1$ equal to $0$ .

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

Step 4.2

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

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$2 \cos (x) = - 1$

Step 4.2.2

Divide each term in $2 \cos (x) = - 1$ by $2$ and simplify.

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Step 4.2.2.1

Divide each term in $2 \cos (x) = - 1$ by $2$ .

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

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

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

Step 4.2.2.2.1.2

Divide $\cos (x)$ by $1$ .

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

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.

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

Step 4.2.3

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

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

Step 4.2.4

Simplify the right side.

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Step 4.2.4.1

The exact value of $\arccos (- \frac{1}{2})$ is $\frac{2 π}{3}$ .

$x = \frac{2 π}{3}$

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

$x = 2 π - \frac{2 π}{3}$

Step 4.2.6

Simplify $2 π - \frac{2 π}{3}$ .

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Step 4.2.6.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 4.2.6.2

Combine fractions.

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Step 4.2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 4.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 4.2.6.3

Simplify the numerator.

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Step 4.2.6.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - 2 π}{3}$

Step 4.2.6.3.2

Subtract $2 π$ from $6 π$ .

$x = \frac{4 π}{3}$

Step 4.2.7

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

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Step 4.2.7.1

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

$\frac{2 π}{| b |}$

Step 4.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 4.2.7.3

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

$\frac{2 π}{1}$

Step 4.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2.8

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

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

Step 5

Set $2 \cos (x) - 1$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $2 \cos (x) - 1$ equal to $0$ .

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

Step 5.2

Solve $2 \cos (x) - 1 = 0$ for $x$ .

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Step 5.2.1

Add $1$ to both sides of the equation.

$2 \cos (x) = 1$

Step 5.2.2

Divide each term in $2 \cos (x) = 1$ by $2$ and simplify.

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Step 5.2.2.1

Divide each term in $2 \cos (x) = 1$ by $2$ .

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

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

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

Step 5.2.2.2.1.2

Divide $\cos (x)$ by $1$ .

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

Step 5.2.3

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

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

Step 5.2.4

Simplify the right side.

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Step 5.2.4.1

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$x = \frac{π}{3}$

Step 5.2.5

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.

$x = 2 π - \frac{π}{3}$

Step 5.2.6

Simplify $2 π - \frac{π}{3}$ .

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Step 5.2.6.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 5.2.6.2

Combine fractions.

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Step 5.2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 5.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 5.2.6.3

Simplify the numerator.

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Step 5.2.6.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 5.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 5.2.7

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

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Step 5.2.7.1

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

$\frac{2 π}{| b |}$

Step 5.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 5.2.7.3

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

$\frac{2 π}{1}$

Step 5.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.8

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

$x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 6

Set $\cos (x) + 1$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $\cos (x) + 1$ equal to $0$ .

$\cos (x) + 1 = 0$

Step 6.2

Solve $\cos (x) + 1 = 0$ for $x$ .

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Step 6.2.1

Subtract $1$ from both sides of the equation.

$\cos (x) = - 1$

Step 6.2.2

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

$x = \arccos (- 1)$

Step 6.2.3

Simplify the right side.

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Step 6.2.3.1

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

$x = π$

Step 6.2.4

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 6.2.5

Subtract $π$ from $2 π$ .

$x = π$

Step 6.2.6

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

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Step 6.2.6.1

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

$\frac{2 π}{| b |}$

Step 6.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 6.2.6.3

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

$\frac{2 π}{1}$

Step 6.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.7

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 7

Set $\cos (x) - 1$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $\cos (x) - 1$ equal to $0$ .

$\cos (x) - 1 = 0$

Step 7.2

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

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Step 7.2.1

Add $1$ to both sides of the equation.

$\cos (x) = 1$

Step 7.2.2

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

$x = \arccos (1)$

Step 7.2.3

Simplify the right side.

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Step 7.2.3.1

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

$x = 0$

Step 7.2.4

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

$x = 2 π - 0$

Step 7.2.5

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 7.2.6

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

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Step 7.2.6.1

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

$\frac{2 π}{| b |}$

Step 7.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 7.2.6.3

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

$\frac{2 π}{1}$

Step 7.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 7.2.7

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 8

The final solution is all the values that make $2 (2 \cos (x) + 1) (2 \cos (x) - 1) (\cos (x) + 1) (\cos (x) - 1) = 0$ true.

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n, π + 2 π n, 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 9

Consolidate the answers.

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