Calculus Examples

$y = \sec^{2} (x)$ , $y = 8 \cos (x)$ , $- \frac{π}{3} \leq x \leq \frac{π}{3}$

Step 1

Solve by substitution to find the intersection between the curves.

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Step 1.1

Eliminate the equal sides of each equation and combine.

$\sec^{2} (x) = 8 \cos (x)$

Step 1.2

Solve $\sec^{2} (x) = 8 \cos (x)$ for $x$ .

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Step 1.2.1

Divide each term in $\sec^{2} (x) = 8 \cos (x)$ by $\sec^{2} (x)$ and simplify.

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Step 1.2.1.1

Divide each term in $\sec^{2} (x) = 8 \cos (x)$ by $\sec^{2} (x)$ .

$\frac{\sec^{2} (x)}{\sec^{2} (x)} = \frac{8 \cos (x)}{\sec^{2} (x)}$

Step 1.2.1.2

Simplify the left side.

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Step 1.2.1.2.1

Cancel the common factor of $\sec^{2} (x)$ .

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Step 1.2.1.2.1.1

Cancel the common factor.

$\frac{\sec^{2} (x)}{\sec^{2} (x)} = \frac{8 \cos (x)}{\sec^{2} (x)}$

Step 1.2.1.2.1.2

Rewrite the expression.

$1 = \frac{8 \cos (x)}{\sec^{2} (x)}$

Step 1.2.1.3

Simplify the right side.

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Step 1.2.1.3.1

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

$1 = \frac{8 \cos (x)}{\sec (x) \sec (x)}$

Step 1.2.1.3.2

Separate fractions.

$1 = \frac{8}{\sec (x)} \cdot \frac{\cos (x)}{\sec (x)}$

Step 1.2.1.3.3

Rewrite $\sec (x)$ in terms of sines and cosines.

$1 = \frac{8}{\sec (x)} \cdot \frac{\cos (x)}{\frac{1}{\cos (x)}}$

Step 1.2.1.3.4

Multiply by the reciprocal of the fraction to divide by $\frac{1}{\cos (x)}$ .

$1 = \frac{8}{\sec (x)} (\cos (x) \cos (x))$

Step 1.2.1.3.5

Simplify.

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Step 1.2.1.3.5.1

Raise $\cos (x)$ to the power of $1$ .

$1 = \frac{8}{\sec (x)} (\cos^{1} (x) \cos (x))$

Step 1.2.1.3.5.2

Raise $\cos (x)$ to the power of $1$ .

$1 = \frac{8}{\sec (x)} (\cos^{1} (x) \cos^{1} (x))$

Step 1.2.1.3.5.3

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

$1 = \frac{8}{\sec (x)} {\cos (x)}^{1 + 1}$

Step 1.2.1.3.5.4

Add $1$ and $1$ .

$1 = \frac{8}{\sec (x)} \cos^{2} (x)$

Step 1.2.1.3.6

Separate fractions.

$1 = \frac{8}{1} \cdot \frac{1}{\sec (x)} \cos^{2} (x)$

Step 1.2.1.3.7

Rewrite $\sec (x)$ in terms of sines and cosines.

$1 = \frac{8}{1} \cdot \frac{1}{\frac{1}{\cos (x)}} \cos^{2} (x)$

Step 1.2.1.3.8

Multiply by the reciprocal of the fraction to divide by $\frac{1}{\cos (x)}$ .

$1 = \frac{8}{1} (1 \cos (x)) \cos^{2} (x)$

Step 1.2.1.3.9

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

$1 = \frac{8}{1} \cos (x) \cos^{2} (x)$

Step 1.2.1.3.10

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

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Step 1.2.1.3.10.1

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

$1 = \frac{8}{1} (\cos^{2} (x) \cos (x))$

Step 1.2.1.3.10.2

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

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Step 1.2.1.3.10.2.1

Raise $\cos (x)$ to the power of $1$ .

$1 = \frac{8}{1} (\cos^{2} (x) \cos^{1} (x))$

Step 1.2.1.3.10.2.2

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

$1 = \frac{8}{1} {\cos (x)}^{2 + 1}$

Step 1.2.1.3.10.3

Add $2$ and $1$ .

$1 = \frac{8}{1} \cos^{3} (x)$

Step 1.2.1.3.11

Divide $8$ by $1$ .

$1 = 8 \cos^{3} (x)$

Step 1.2.2

Rewrite the equation as $8 \cos^{3} (x) = 1$ .

$8 \cos^{3} (x) = 1$

Step 1.2.3

Subtract $1$ from both sides of the equation.

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

Step 1.2.4

Factor the left side of the equation.

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Step 1.2.4.1

Rewrite $8 \cos^{3} (x)$ as ${(2 \cos (x))}^{3}$ .

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

Step 1.2.4.2

Rewrite $1$ as $1^{3}$ .

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

Step 1.2.4.3

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = 2 \cos (x)$ and $b = 1$ .

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

Step 1.2.4.4

Simplify.

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Step 1.2.4.4.1

Apply the product rule to $2 \cos (x)$ .

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

Step 1.2.4.4.2

Raise $2$ to the power of $2$ .

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

Step 1.2.4.4.3

Multiply $2$ by $1$ .

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

Step 1.2.4.4.4

One to any power is one.

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

Step 1.2.5

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

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

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

Step 1.2.6

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

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Step 1.2.6.1

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

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

Step 1.2.6.2

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

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Step 1.2.6.2.1

Add $1$ to both sides of the equation.

$2 \cos (x) = 1$

Step 1.2.6.2.2

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

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Step 1.2.6.2.2.1

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

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

Step 1.2.6.2.2.2

Simplify the left side.

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Step 1.2.6.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.6.2.2.2.1.1

Cancel the common factor.

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

Step 1.2.6.2.2.2.1.2

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

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

Step 1.2.7

Set $4 \cos^{2} (x) + 2 \cos (x) + 1$ equal to $0$ and solve for $\cos (x)$ .

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Step 1.2.7.1

Set $4 \cos^{2} (x) + 2 \cos (x) + 1$ equal to $0$ .

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

Step 1.2.7.2

Solve $4 \cos^{2} (x) + 2 \cos (x) + 1 = 0$ for $\cos (x)$ .

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Step 1.2.7.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a} + y = 8 \cos (x)$

Step 1.2.7.2.2

Substitute the values $a = 4$ , $b = 2$ , and $c = 1$ into the quadratic formula and solve for $\cos (x)$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (4 \cdot 1)}}{2 \cdot 4} + y = 8 \cos (x)$

Step 1.2.7.2.3

Simplify.

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Step 1.2.7.2.3.1

Simplify the numerator.

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Step 1.2.7.2.3.1.1

Raise $2$ to the power of $2$ .

$\cos (x) = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}$

Step 1.2.7.2.3.1.2

Multiply $- 4 \cdot 4 \cdot 1$ .

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Step 1.2.7.2.3.1.2.1

Multiply $- 4$ by $4$ .

$\cos (x) = \frac{- 2 \pm \sqrt{4 - 16 \cdot 1}}{2 \cdot 4}$

Step 1.2.7.2.3.1.2.2

Multiply $- 16$ by $1$ .

$\cos (x) = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

Step 1.2.7.2.3.1.3

Subtract $16$ from $4$ .

$\cos (x) = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 4}$

Step 1.2.7.2.3.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$\cos (x) = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

Step 1.2.7.2.3.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$\cos (x) = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

Step 1.2.7.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$\cos (x) = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

Step 1.2.7.2.3.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.2.7.2.3.1.7.1

Factor $4$ out of $12$ .

$\cos (x) = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

Step 1.2.7.2.3.1.7.2

Rewrite $4$ as $2^{2}$ .

$\cos (x) = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

Step 1.2.7.2.3.1.8

Pull terms out from under the radical.

$\cos (x) = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

Step 1.2.7.2.3.1.9

Move $2$ to the left of $i$ .

$\cos (x) = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

Step 1.2.7.2.3.2

Multiply $2$ by $4$ .

$\cos (x) = \frac{- 2 \pm 2 i \sqrt{3}}{8}$

Step 1.2.7.2.3.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{8}$ .

$\cos (x) = \frac{- 1 \pm i \sqrt{3}}{4}$

Step 1.2.7.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 1.2.7.2.4.1

Simplify the numerator.

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Step 1.2.7.2.4.1.1

Raise $2$ to the power of $2$ .

$\cos (x) = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}$

Step 1.2.7.2.4.1.2

Multiply $- 4 \cdot 4 \cdot 1$ .

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Step 1.2.7.2.4.1.2.1

Multiply $- 4$ by $4$ .

$\cos (x) = \frac{- 2 \pm \sqrt{4 - 16 \cdot 1}}{2 \cdot 4}$

Step 1.2.7.2.4.1.2.2

Multiply $- 16$ by $1$ .

$\cos (x) = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

Step 1.2.7.2.4.1.3

Subtract $16$ from $4$ .

$\cos (x) = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 4}$

Step 1.2.7.2.4.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$\cos (x) = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

Step 1.2.7.2.4.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$\cos (x) = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

Step 1.2.7.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$\cos (x) = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

Step 1.2.7.2.4.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.2.7.2.4.1.7.1

Factor $4$ out of $12$ .

$\cos (x) = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

Step 1.2.7.2.4.1.7.2

Rewrite $4$ as $2^{2}$ .

$\cos (x) = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

Step 1.2.7.2.4.1.8

Pull terms out from under the radical.

$\cos (x) = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

Step 1.2.7.2.4.1.9

Move $2$ to the left of $i$ .

$\cos (x) = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

Step 1.2.7.2.4.2

Multiply $2$ by $4$ .

$\cos (x) = \frac{- 2 \pm 2 i \sqrt{3}}{8}$

Step 1.2.7.2.4.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{8}$ .

$\cos (x) = \frac{- 1 \pm i \sqrt{3}}{4}$

Step 1.2.7.2.4.4

Change the $\pm$ to $+$ .

$\cos (x) = \frac{- 1 + i \sqrt{3}}{4}$

Step 1.2.7.2.4.5

Rewrite $- 1$ as $- 1 (1)$ .

$\cos (x) = \frac{- 1 \cdot 1 + i \sqrt{3}}{4}$

Step 1.2.7.2.4.6

Factor $- 1$ out of $i \sqrt{3}$ .

$\cos (x) = \frac{- 1 \cdot 1 - (- i \sqrt{3})}{4}$

Step 1.2.7.2.4.7

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{3})$ .

$\cos (x) = \frac{- 1 (1 - i \sqrt{3})}{4}$

Step 1.2.7.2.4.8

Move the negative in front of the fraction.

$\cos (x) = - \frac{1 - i \sqrt{3}}{4}$

Step 1.2.7.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 1.2.7.2.5.1

Simplify the numerator.

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Step 1.2.7.2.5.1.1

Raise $2$ to the power of $2$ .

$\cos (x) = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}$

Step 1.2.7.2.5.1.2

Multiply $- 4 \cdot 4 \cdot 1$ .

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Step 1.2.7.2.5.1.2.1

Multiply $- 4$ by $4$ .

$\cos (x) = \frac{- 2 \pm \sqrt{4 - 16 \cdot 1}}{2 \cdot 4}$

Step 1.2.7.2.5.1.2.2

Multiply $- 16$ by $1$ .

$\cos (x) = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

Step 1.2.7.2.5.1.3

Subtract $16$ from $4$ .

$\cos (x) = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 4}$

Step 1.2.7.2.5.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$\cos (x) = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

Step 1.2.7.2.5.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$\cos (x) = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

Step 1.2.7.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$\cos (x) = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

Step 1.2.7.2.5.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.2.7.2.5.1.7.1

Factor $4$ out of $12$ .

$\cos (x) = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

Step 1.2.7.2.5.1.7.2

Rewrite $4$ as $2^{2}$ .

$\cos (x) = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

Step 1.2.7.2.5.1.8

Pull terms out from under the radical.

$\cos (x) = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

Step 1.2.7.2.5.1.9

Move $2$ to the left of $i$ .

$\cos (x) = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

Step 1.2.7.2.5.2

Multiply $2$ by $4$ .

$\cos (x) = \frac{- 2 \pm 2 i \sqrt{3}}{8}$

Step 1.2.7.2.5.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{8}$ .

$\cos (x) = \frac{- 1 \pm i \sqrt{3}}{4}$

Step 1.2.7.2.5.4

Change the $\pm$ to $-$ .

$\cos (x) = \frac{- 1 - i \sqrt{3}}{4}$

Step 1.2.7.2.5.5

Rewrite $- 1$ as $- 1 (1)$ .

$\cos (x) = \frac{- 1 \cdot 1 - i \sqrt{3}}{4}$

Step 1.2.7.2.5.6

Factor $- 1$ out of $- i \sqrt{3}$ .

$\cos (x) = \frac{- 1 \cdot 1 - (i \sqrt{3})}{4}$

Step 1.2.7.2.5.7

Factor $- 1$ out of $- 1 (1) - (i \sqrt{3})$ .

$\cos (x) = \frac{- 1 (1 + i \sqrt{3})}{4}$

Step 1.2.7.2.5.8

Move the negative in front of the fraction.

$\cos (x) = - \frac{1 + i \sqrt{3}}{4}$

Step 1.2.7.2.6

The final answer is the combination of both solutions.

$\cos (x) = - \frac{1 - i \sqrt{3}}{4}, - \frac{1 + i \sqrt{3}}{4}$

Step 1.2.8

The final solution is all the values that make $(2 \cos (x) - 1) (4 \cos^{2} (x) + 2 \cos (x) + 1) = 0$ true.

$\cos (x) = \frac{1}{2}, - \frac{1 - i \sqrt{3}}{4}, - \frac{1 + i \sqrt{3}}{4}$

Step 1.2.9

Set up each of the solutions to solve for $x$ .

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

$\cos (x) = - \frac{1 - i \sqrt{3}}{4}$

$\cos (x) = - \frac{1 + i \sqrt{3}}{4} + y = 8 \cos (x)$

Step 1.2.10

Solve for $x$ in $\cos (x) = \frac{1}{2}$ .

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Step 1.2.10.1

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

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

Step 1.2.10.2

Simplify the right side.

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Step 1.2.10.2.1

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

$x = \frac{π}{3}$

Step 1.2.10.3

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.

$x = 2 π - \frac{π}{3}$

Step 1.2.10.4

Simplify $2 π - \frac{π}{3}$ .

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Step 1.2.10.4.1

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 1.2.10.4.2

Combine fractions.

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Step 1.2.10.4.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 1.2.10.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 1.2.10.4.3

Simplify the numerator.

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Step 1.2.10.4.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 1.2.10.4.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 1.2.10.5

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

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Step 1.2.10.5.1

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

$\frac{2 π}{| b |}$

Step 1.2.10.5.2

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

$\frac{2 π}{| 1 |}$

Step 1.2.10.5.3

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

$\frac{2 π}{1}$

Step 1.2.10.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.10.6

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

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

Step 1.2.11

Solve for $x$ in $\cos (x) = - \frac{1 - i \sqrt{3}}{4}$ .

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Step 1.2.11.1

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

$x = \arccos (- \frac{1 - i \sqrt{3}}{4})$

Step 1.2.11.2

The inverse cosine of $\arccos (- \frac{1 - i \sqrt{3}}{4})$ is undefined.

$Undefined + y = 8 \cos (x)$

Step 1.2.12

Solve for $x$ in $\cos (x) = - \frac{1 + i \sqrt{3}}{4}$ .

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Step 1.2.12.1

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

$x = \arccos (- \frac{1 + i \sqrt{3}}{4})$

Step 1.2.12.2

The inverse cosine of $\arccos (- \frac{1 + i \sqrt{3}}{4})$ is undefined.

$Undefined + y = 8 \cos (x)$

Step 1.2.13

List all of the solutions.

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

Step 1.3

Evaluate $y$ when $x = \frac{π}{3} + 2 π n$ .

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Step 1.3.1

Substitute $\frac{π}{3} + 2 π n$ for $x$ .

$y = 8 \cos (\frac{π}{3} + 2 π n)$

Step 1.3.2

Remove parentheses.

$y = 8 \cos (\frac{π}{3} + 2 π n)$

Step 1.4

Evaluate $y$ when $x = \frac{5 π}{3} + 2 π n$ .

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Step 1.4.1

Substitute $\frac{5 π}{3} + 2 π n$ for $x$ .

$y = 8 \cos (\frac{5 π}{3} + 2 π n)$

Step 1.4.2

Remove parentheses.

$y = 8 \cos (\frac{5 π}{3} + 2 π n)$

Step 1.5

List all of the solutions.

$y = 8 \cos (\frac{π}{3} + 2 π n), x = \frac{π}{3} + 2 π n$

$y = 8 \cos (\frac{5 π}{3} + 2 π n), x = \frac{5 π}{3} + 2 π n$

$y = 8 \cos (\frac{π}{3} + 2 π n), x = \frac{π}{3} + 2 π n$

$y = 8 \cos (\frac{5 π}{3} + 2 π n), x = \frac{5 π}{3} + 2 π n$

Step 2

The area of the region between the curves is defined as the integral of the upper curve minus the integral of the lower curve over each region. The regions are determined by the intersection points of the curves. This can be done algebraically or graphically.

$A r e a = \int_{- \frac{π}{3}}^{\frac{π}{3}} 8 \cos (x) d x - \int_{- \frac{π}{3}}^{\frac{π}{3}} \sec^{2} (x) d x$

Step 3

Integrate to find the area between $- \frac{π}{3}$ and $\frac{π}{3}$ .

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Step 3.1

Combine the integrals into a single integral.

$\int_{- \frac{π}{3}}^{\frac{π}{3}} 8 \cos (x) - (\sec^{2} (x)) d x$

Step 3.2

Simplify each term.

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Step 3.2.1

Rewrite $\sec (x)$ in terms of sines and cosines.

$8 \cos (x) - {(\frac{1}{\cos (x)})}^{2}$

Step 3.2.2

Apply the product rule to $\frac{1}{\cos (x)}$ .

$8 \cos (x) - \frac{1^{2}}{\cos^{2} (x)}$

Step 3.2.3

One to any power is one.

$8 \cos (x) - \frac{1}{\cos^{2} (x)}$

$\int_{- \frac{π}{3}}^{\frac{π}{3}} 8 \cos (x) - \frac{1}{\cos^{2} (x)} d x$

Step 3.3

Simplify each term.

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Step 3.3.1

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

$8 \cos (x) - \frac{1^{2}}{\cos^{2} (x)}$

Step 3.3.2

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

$8 \cos (x) - {(\frac{1}{\cos (x)})}^{2}$

Step 3.3.3

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

$8 \cos (x) - \sec^{2} (x)$

$\int_{- \frac{π}{3}}^{\frac{π}{3}} 8 \cos (x) - \sec^{2} (x) d x$

Step 3.4

Split the single integral into multiple integrals.

$\int_{- \frac{π}{3}}^{\frac{π}{3}} 8 \cos (x) d x + \int_{- \frac{π}{3}}^{\frac{π}{3}} - \sec^{2} (x) d x$

Step 3.5

Since $8$ is constant with respect to $x$ , move $8$ out of the integral.

$8 \int_{- \frac{π}{3}}^{\frac{π}{3}} \cos (x) d x + \int_{- \frac{π}{3}}^{\frac{π}{3}} - \sec^{2} (x) d x$

Step 3.6

The integral of $\cos (x)$ with respect to $x$ is $\sin (x)$ .

$8 (\sin (x)]_{- \frac{π}{3}}^{\frac{π}{3}}) + \int_{- \frac{π}{3}}^{\frac{π}{3}} - \sec^{2} (x) d x$

Step 3.7

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$8 (\sin (x)]_{- \frac{π}{3}}^{\frac{π}{3}}) - \int_{- \frac{π}{3}}^{\frac{π}{3}} \sec^{2} (x) d x$

Step 3.8

Since the derivative of $\tan (x)$ is $\sec^{2} (x)$ , the integral of $\sec^{2} (x)$ is $\tan (x)$ .

$8 (\sin (x)]_{- \frac{π}{3}}^{\frac{π}{3}}) - (\tan (x)]_{- \frac{π}{3}}^{\frac{π}{3}})$

Step 3.9

Simplify the answer.

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Step 3.9.1

Substitute and simplify.

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Step 3.9.1.1

Evaluate $\sin (x)$ at $\frac{π}{3}$ and at $- \frac{π}{3}$ .

$8 ((\sin (\frac{π}{3})) - \sin (- \frac{π}{3})) - (\tan (x)]_{- \frac{π}{3}}^{\frac{π}{3}})$

Step 3.9.1.2

Evaluate $\tan (x)$ at $\frac{π}{3}$ and at $- \frac{π}{3}$ .

$8 (\sin (\frac{π}{3}) - \sin (- \frac{π}{3})) - (\tan (\frac{π}{3}) - \tan (- \frac{π}{3}))$

Step 3.9.2

Simplify.

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Step 3.9.2.1

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

$8 (\frac{\sqrt{3}}{2} - \sin (- \frac{π}{3})) - (\tan (\frac{π}{3}) - \tan (- \frac{π}{3}))$

Step 3.9.2.2

The exact value of $\tan (\frac{π}{3})$ is $\sqrt{3}$ .

$8 (\frac{\sqrt{3}}{2} - \sin (- \frac{π}{3})) - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3

Simplify.

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Step 3.9.3.1

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$8 (\frac{\sqrt{3}}{2} - \sin (\frac{5 π}{3})) - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$8 (\frac{\sqrt{3}}{2} - - \sin (\frac{π}{3})) - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3.3

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

$8 (\frac{\sqrt{3}}{2} - - \frac{\sqrt{3}}{2}) - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3.4

Multiply $- - \frac{\sqrt{3}}{2}$ .

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Step 3.9.3.4.1

Multiply $- 1$ by $- 1$ .

$8 (\frac{\sqrt{3}}{2} + 1 \frac{\sqrt{3}}{2}) - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3.4.2

Multiply $\frac{\sqrt{3}}{2}$ by $1$ .

$8 (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3.5

Combine the numerators over the common denominator.

$8 \frac{\sqrt{3} + \sqrt{3}}{2} - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3.6

Add $\sqrt{3}$ and $\sqrt{3}$ .

$8 \frac{2 \sqrt{3}}{2} - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3.7

Cancel the common factor of $2$ .

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Step 3.9.3.7.1

Factor $2$ out of $8$ .

$2 (4) \frac{2 \sqrt{3}}{2} - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3.7.2

Cancel the common factor.

$2 \cdot 4 \frac{2 \sqrt{3}}{2} - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3.7.3

Rewrite the expression.

$4 (2 \sqrt{3}) - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3.8

Multiply $2$ by $4$ .

$8 \sqrt{3} - (\sqrt{3} - \tan (- \frac{π}{3}))$

Step 3.9.3.9

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$8 \sqrt{3} - (\sqrt{3} - \tan (\frac{5 π}{3}))$

Step 3.9.3.10

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the fourth quadrant.

$8 \sqrt{3} - (\sqrt{3} - - \tan (\frac{π}{3}))$

Step 3.9.3.11

The exact value of $\tan (\frac{π}{3})$ is $\sqrt{3}$ .

$8 \sqrt{3} - (\sqrt{3} - - \sqrt{3})$

Step 3.9.3.12

Multiply $- - \sqrt{3}$ .

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Step 3.9.3.12.1

Multiply $- 1$ by $- 1$ .

$8 \sqrt{3} - (\sqrt{3} + 1 \sqrt{3})$

Step 3.9.3.12.2

Multiply $\sqrt{3}$ by $1$ .

$8 \sqrt{3} - (\sqrt{3} + \sqrt{3})$

Step 3.9.3.13

Add $\sqrt{3}$ and $\sqrt{3}$ .

$8 \sqrt{3} - (2 \sqrt{3})$

Step 3.9.3.14

Multiply $2$ by $- 1$ .

$8 \sqrt{3} - 2 \sqrt{3}$

Step 3.9.3.15

Subtract $2 \sqrt{3}$ from $8 \sqrt{3}$ .

$6 \sqrt{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$6 \sqrt{3}$

Decimal Form:

$10.39230484 \dots$

Step 5