Precalculus Examples

$5 \cos^{3} (x) = 5 \cos (x)$

Step 1

Subtract $5 \cos (x)$ from both sides of the equation.

$5 \cos^{3} (x) - 5 \cos (x) = 0$

Step 2

Factor $5 \cos^{3} (x) - 5 \cos (x)$ .

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Step 2.1

Factor $5 \cos (x)$ out of $5 \cos^{3} (x) - 5 \cos (x)$ .

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Step 2.1.1

Factor $5 \cos (x)$ out of $5 \cos^{3} (x)$ .

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

Step 2.1.2

Factor $5 \cos (x)$ out of $- 5 \cos (x)$ .

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

Step 2.1.3

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

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

Step 2.2

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

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

Step 2.3

Factor.

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Step 2.3.1

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 $b = 1$ .

$5 \cos (x) ((\cos (x) + 1) (\cos (x) - 1)) = 0$

Step 2.3.2

Remove unnecessary parentheses.

$5 \cos (x) (\cos (x) + 1) (\cos (x) - 1) = 0$

Step 3

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

$\cos (x) = 0$

$\cos (x) + 1 = 0$

$\cos (x) - 1 = 0$

Step 4

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

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Step 4.1

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

$\cos (x) = 0$

Step 4.2

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

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Step 4.2.1

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

$x = \arccos (0)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 4.2.3

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.

$x = 2 π - \frac{π}{2}$

Step 4.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 4.2.4.1

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

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

Step 4.2.4.2

Combine fractions.

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Step 4.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 4.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 4.2.4.3

Simplify the numerator.

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Step 4.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 4.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 4.2.5

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

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Step 4.2.5.1

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

$\frac{2 π}{| b |}$

Step 4.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 4.2.5.3

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

$\frac{2 π}{1}$

Step 4.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2.6

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

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

Step 5

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

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Step 5.1

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

$\cos (x) + 1 = 0$

Step 5.2

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

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Step 5.2.1

Subtract $1$ from both sides of the equation.

$\cos (x) = - 1$

Step 5.2.2

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

$x = \arccos (- 1)$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

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

$x = π$

Step 5.2.4

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 5.2.5

Subtract $π$ from $2 π$ .

$x = π$

Step 5.2.6

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

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Step 5.2.6.1

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

$\frac{2 π}{| b |}$

Step 5.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 5.2.6.3

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

$\frac{2 π}{1}$

Step 5.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.7

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 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

Add $1$ to 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 $0$ .

$x = 0$

Step 6.2.4

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 6.2.5

Subtract $0$ from $2 π$ .

$x = 2 π$

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.

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

Step 7

The final solution is all the values that make $5 \cos (x) (\cos (x) + 1) (\cos (x) - 1) = 0$ true.

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

Step 8

Consolidate the answers.

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