Calculus Examples

$2 \cos (3 x) + 1 = \cos (2 x) + 2 \cos (x)$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $\cos (2 x)$ from both sides of the equation.

$2 \cos (3 x) + 1 - \cos (2 x) = 2 \cos (x)$

Step 1.2

Subtract $2 \cos (x)$ from both sides of the equation.

$2 \cos (3 x) + 1 - \cos (2 x) - 2 \cos (x) = 0$

Step 2

Simplify the left side of the equation.

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Step 2.1

Simplify each term.

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Step 2.1.1

Use the triple-angle identity to transform $\cos (3 x)$ to $4 \cos^{3} (x) - 3 \cos (x)$ .

$2 (4 \cos^{3} (x) - 3 \cos (x)) + 1 - \cos (2 x) - 2 \cos (x) = 0$

Step 2.1.2

Apply the distributive property.

$2 (4 \cos^{3} (x)) + 2 (- 3 \cos (x)) + 1 - \cos (2 x) - 2 \cos (x) = 0$

Step 2.1.3

Multiply $4$ by $2$ .

$8 \cos^{3} (x) + 2 (- 3 \cos (x)) + 1 - \cos (2 x) - 2 \cos (x) = 0$

Step 2.1.4

Multiply $- 3$ by $2$ .

$8 \cos^{3} (x) - 6 \cos (x) + 1 - \cos (2 x) - 2 \cos (x) = 0$

Step 2.1.5

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

$8 \cos^{3} (x) - 6 \cos (x) + 1 - (2 \cos^{2} (x) - 1) - 2 \cos (x) = 0$

Step 2.1.6

Apply the distributive property.

$8 \cos^{3} (x) - 6 \cos (x) + 1 - (2 \cos^{2} (x)) + 1 - 2 \cos (x) = 0$

Step 2.1.7

Multiply $2$ by $- 1$ .

$8 \cos^{3} (x) - 6 \cos (x) + 1 - 2 \cos^{2} (x) + 1 - 2 \cos (x) = 0$

Step 2.1.8

Multiply $- 1$ by $- 1$ .

$8 \cos^{3} (x) - 6 \cos (x) + 1 - 2 \cos^{2} (x) + 1 - 2 \cos (x) = 0$

Step 2.2

Simplify by adding terms.

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Step 2.2.1

Subtract $2 \cos (x)$ from $- 6 \cos (x)$ .

$8 \cos^{3} (x) + 1 - 2 \cos^{2} (x) + 1 - 8 \cos (x) = 0$

Step 2.2.2

Add $1$ and $1$ .

$8 \cos^{3} (x) + 2 - 2 \cos^{2} (x) - 8 \cos (x) = 0$

Step 3

Factor $8 \cos^{3} (x) + 2 - 2 \cos^{2} (x) - 8 \cos (x)$ .

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Step 3.1

Factor $2$ out of $8 \cos^{3} (x) + 2 - 2 \cos^{2} (x) - 8 \cos (x)$ .

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Step 3.1.1

Factor $2$ out of $8 \cos^{3} (x)$ .

$2 (4 \cos^{3} (x)) + 2 - 2 \cos^{2} (x) - 8 \cos (x) = 0$

Step 3.1.2

Factor $2$ out of $2$ .

$2 (4 \cos^{3} (x)) + 2 (1) - 2 \cos^{2} (x) - 8 \cos (x) = 0$

Step 3.1.3

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

$2 (4 \cos^{3} (x)) + 2 (1) + 2 (- \cos^{2} (x)) - 8 \cos (x) = 0$

Step 3.1.4

Factor $2$ out of $- 8 \cos (x)$ .

$2 (4 \cos^{3} (x)) + 2 (1) + 2 (- \cos^{2} (x)) + 2 (- 4 \cos (x)) = 0$

Step 3.1.5

Factor $2$ out of $2 (4 \cos^{3} (x)) + 2 (1)$ .

$2 (4 \cos^{3} (x) + 1) + 2 (- \cos^{2} (x)) + 2 (- 4 \cos (x)) = 0$

Step 3.1.6

Factor $2$ out of $2 (4 \cos^{3} (x) + 1) + 2 (- \cos^{2} (x))$ .

$2 (4 \cos^{3} (x) + 1 - \cos^{2} (x)) + 2 (- 4 \cos (x)) = 0$

Step 3.1.7

Factor $2$ out of $2 (4 \cos^{3} (x) + 1 - \cos^{2} (x)) + 2 (- 4 \cos (x))$ .

$2 (4 \cos^{3} (x) + 1 - \cos^{2} (x) - 4 \cos (x)) = 0$

Step 3.2

Reorder terms.

$2 (4 \cos^{3} (x) - \cos^{2} (x) - 4 \cos (x) + 1) = 0$

Step 3.3

Factor out the greatest common factor from each group.

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Step 3.3.1

Group the first two terms and the last two terms.

$2 (4 \cos^{3} (x) - \cos^{2} (x) - 4 \cos (x) + 1) = 0$

Step 3.3.2

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

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

Factor.

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Step 3.6.1

Factor.

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Step 3.6.1.1

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 $b = 1$ .

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

Step 3.6.1.2

Remove unnecessary parentheses.

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

Step 3.6.2

Remove unnecessary parentheses.

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

Step 4

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

$4 \cos (x) - 1 = 0$

$\cos (x) + 1 = 0$

$\cos (x) - 1 = 0$

Step 5

Set $4 \cos (x) - 1$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $4 \cos (x) - 1$ equal to $0$ .

$4 \cos (x) - 1 = 0$

Step 5.2

Solve $4 \cos (x) - 1 = 0$ for $x$ .

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Step 5.2.1

Add $1$ to both sides of the equation.

$4 \cos (x) = 1$

Step 5.2.2

Divide each term in $4 \cos (x) = 1$ by $4$ and simplify.

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Step 5.2.2.1

Divide each term in $4 \cos (x) = 1$ by $4$ .

$\frac{4 \cos (x)}{4} = \frac{1}{4}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $4$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{4 \cos (x)}{4} = \frac{1}{4}$

Step 5.2.2.2.1.2

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

$\cos (x) = \frac{1}{4}$

Step 5.2.3

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

$x = \arccos (\frac{1}{4})$

Step 5.2.4

Simplify the right side.

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Step 5.2.4.1

Evaluate $\arccos (\frac{1}{4})$ .

$x = 1.31811607$

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.

$x = 2 (3.14159265) - 1.31811607$

Step 5.2.6

Solve for $x$ .

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Step 5.2.6.1

Remove parentheses.

$x = 2 (3.14159265) - 1.31811607$

Step 5.2.6.2

Simplify $2 (3.14159265) - 1.31811607$ .

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Step 5.2.6.2.1

Multiply $2$ by $3.14159265$ .

$x = 6.2831853 - 1.31811607$

Step 5.2.6.2.2

Subtract $1.31811607$ from $6.2831853$ .

$x = 4.96506923$

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.

$x = 1.31811607 + 2 π n, 4.96506923 + 2 π n$ , 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 $2 (4 \cos (x) - 1) (\cos (x) + 1) (\cos (x) - 1) = 0$ true.

$x = 1.31811607 + 2 π n, 4.96506923 + 2 π n, π + 2 π n, 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 9

Consolidate the answers.

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Step 9.1

Consolidate $π + 2 π n$ and $2 π n$ to $π n$ .

$x = 1.31811607 + 2 π n, 4.96506923 + 2 π n, π n, 2 π + 2 π n$ , for any integer $n$

Step 9.2

Consolidate $π n$ and $2 π + 2 π n$ to $π n$ .

$x = 1.31811607 + 2 π n, 4.96506923 + 2 π n, π n$ , for any integer $n$