Precalculus Examples

$9 (1 - \cos (x)) = \sin^{2} (x)$

Step 1

Divide each term in the equation by $\cos (x)$ .

$\frac{9 (1 - \cos (x))}{\cos (x)} = \frac{\sin^{2} (x)}{\cos (x)}$

Step 2

Replace $\cos (x)$ with an equivalent expression in the numerator.

$9 (1 - \cos (x)) \cdot \sec (x) = \frac{\sin^{2} (x)}{\cos (x)}$

Step 3

Apply the distributive property.

$(9 \cdot 1 + 9 (- \cos (x))) \cdot \sec (x) = \frac{\sin^{2} (x)}{\cos (x)}$

Step 4

Multiply.

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Step 4.1

Multiply $9$ by $1$ .

$(9 + 9 (- \cos (x))) \cdot \sec (x) = \frac{\sin^{2} (x)}{\cos (x)}$

Step 4.2

Multiply $- 1$ by $9$ .

$(9 - 9 \cos (x)) \cdot \sec (x) = \frac{\sin^{2} (x)}{\cos (x)}$

Step 5

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

$(9 - 9 \cos (x)) \cdot \frac{1}{\cos (x)} = \frac{\sin^{2} (x)}{\cos (x)}$

Step 6

Simplify terms.

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Step 6.1

Apply the distributive property.

$9 (\frac{1}{\cos (x)}) - 9 \cos (x) \frac{1}{\cos (x)} = \frac{\sin^{2} (x)}{\cos (x)}$

Step 6.2

Combine $9$ and $\frac{1}{\cos (x)}$ .

$\frac{9}{\cos (x)} - 9 \cos (x) \frac{1}{\cos (x)} = \frac{\sin^{2} (x)}{\cos (x)}$

Step 6.3

Cancel the common factor of $\cos (x)$ .

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Step 6.3.1

Factor $\cos (x)$ out of $- 9 \cos (x)$ .

$\frac{9}{\cos (x)} + \cos (x) \cdot (- 9 \frac{1}{\cos (x)}) = \frac{\sin^{2} (x)}{\cos (x)}$

Step 6.3.2

Cancel the common factor.

$\frac{9}{\cos (x)} + \cos (x) \cdot (- 9 \frac{1}{\cos (x)}) = \frac{\sin^{2} (x)}{\cos (x)}$

Step 6.3.3

Rewrite the expression.

$\frac{9}{\cos (x)} - 9 = \frac{\sin^{2} (x)}{\cos (x)}$

Step 7

Simplify each term.

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Step 7.1

Separate fractions.

$\frac{9}{1} \cdot \frac{1}{\cos (x)} - 9 = \frac{\sin^{2} (x)}{\cos (x)}$

Step 7.2

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

$\frac{9}{1} \cdot \sec (x) - 9 = \frac{\sin^{2} (x)}{\cos (x)}$

Step 7.3

Divide $9$ by $1$ .

$9 \sec (x) - 9 = \frac{\sin^{2} (x)}{\cos (x)}$

Step 8

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

$9 \sec (x) - 9 = \frac{\sin (x) \sin (x)}{\cos (x)}$

Step 9

Separate fractions.

$9 \sec (x) - 9 = \frac{\sin (x)}{1} \cdot \frac{\sin (x)}{\cos (x)}$

Step 10

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$9 \sec (x) - 9 = \frac{\sin (x)}{1} \cdot \tan (x)$

Step 11

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

$9 \sec (x) - 9 = \sin (x) \tan (x)$

Step 12

Simplify the left side.

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Step 12.1

Simplify each term.

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Step 12.1.1

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

$9 \frac{1}{\cos (x)} - 9 = \sin (x) \tan (x)$

Step 12.1.2

Combine $9$ and $\frac{1}{\cos (x)}$ .

$\frac{9}{\cos (x)} - 9 = \sin (x) \tan (x)$

Step 13

Simplify the right side.

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Step 13.1

Simplify $\sin (x) \tan (x)$ .

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Step 13.1.1

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

$\frac{9}{\cos (x)} - 9 = \sin (x) \frac{\sin (x)}{\cos (x)}$

Step 13.1.2

Multiply $\sin (x) \frac{\sin (x)}{\cos (x)}$ .

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Step 13.1.2.1

Combine $\sin (x)$ and $\frac{\sin (x)}{\cos (x)}$ .

$\frac{9}{\cos (x)} - 9 = \frac{\sin (x) \sin (x)}{\cos (x)}$

Step 13.1.2.2

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

$\frac{9}{\cos (x)} - 9 = \frac{\sin^{1} (x) \sin (x)}{\cos (x)}$

Step 13.1.2.3

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

$\frac{9}{\cos (x)} - 9 = \frac{\sin^{1} (x) \sin^{1} (x)}{\cos (x)}$

Step 13.1.2.4

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

$\frac{9}{\cos (x)} - 9 = \frac{{\sin (x)}^{1 + 1}}{\cos (x)}$

Step 13.1.2.5

Add $1$ and $1$ .

$\frac{9}{\cos (x)} - 9 = \frac{\sin^{2} (x)}{\cos (x)}$

Step 14

Multiply both sides of the equation by $\cos (x)$ .

$\cos (x) (\frac{9}{\cos (x)} - 9) = \cos (x) \frac{\sin^{2} (x)}{\cos (x)}$

Step 15

Apply the distributive property.

$\cos (x) \frac{9}{\cos (x)} + \cos (x) \cdot - 9 = \cos (x) \frac{\sin^{2} (x)}{\cos (x)}$

Step 16

Cancel the common factor of $\cos (x)$ .

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Step 16.1

Cancel the common factor.

$\cos (x) \frac{9}{\cos (x)} + \cos (x) \cdot - 9 = \cos (x) \frac{\sin^{2} (x)}{\cos (x)}$

Step 16.2

Rewrite the expression.

$9 + \cos (x) \cdot - 9 = \cos (x) \frac{\sin^{2} (x)}{\cos (x)}$

Step 17

Move $- 9$ to the left of $\cos (x)$ .

$9 - 9 \cdot \cos (x) = \cos (x) \frac{\sin^{2} (x)}{\cos (x)}$

Step 18

Cancel the common factor of $\cos (x)$ .

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Step 18.1

Cancel the common factor.

$9 - 9 \cos (x) = \cos (x) \frac{\sin^{2} (x)}{\cos (x)}$

Step 18.2

Rewrite the expression.

$9 - 9 \cos (x) = \sin^{2} (x)$

Step 19

Subtract $\sin^{2} (x)$ from both sides of the equation.

$9 - 9 \cos (x) - \sin^{2} (x) = 0$

Step 20

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

$9 - 9 \cos (x) - (1 - \cos^{2} (x)) = 0$

Step 21

Solve for $x$ .

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Step 21.1

Substitute $u$ for $\cos (x)$ .

$9 - 9 u - (1 - {(u)}^{2}) = 0$

Step 21.2

Simplify $9 - 9 u - (1 - u^{2})$ .

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Step 21.2.1

Simplify each term.

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Step 21.2.1.1

Apply the distributive property.

$9 - 9 u - 1 \cdot 1 - - u^{2} = 0$

Step 21.2.1.2

Multiply $- 1$ by $1$ .

$9 - 9 u - 1 - - u^{2} = 0$

Step 21.2.1.3

Multiply $- - u^{2}$ .

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Step 21.2.1.3.1

Multiply $- 1$ by $- 1$ .

$9 - 9 u - 1 + 1 u^{2} = 0$

Step 21.2.1.3.2

Multiply $u^{2}$ by $1$ .

$9 - 9 u - 1 + u^{2} = 0$

Step 21.2.2

Subtract $1$ from $9$ .

$- 9 u + 8 + u^{2} = 0$

Step 21.3

Factor $- 9 u + 8 + u^{2}$ using the AC method.

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Step 21.3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $8$ and whose sum is $- 9$ .

$- 8, - 1$

Step 21.3.2

Write the factored form using these integers.

$(u - 8) (u - 1) = 0$

Step 21.4

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

$u - 8 = 0$

$u - 1 = 0$

Step 21.5

Set $u - 8$ equal to $0$ and solve for $u$ .

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Step 21.5.1

Set $u - 8$ equal to $0$ .

$u - 8 = 0$

Step 21.5.2

Add $8$ to both sides of the equation.

$u = 8$

Step 21.6

Set $u - 1$ equal to $0$ and solve for $u$ .

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Step 21.6.1

Set $u - 1$ equal to $0$ .

$u - 1 = 0$

Step 21.6.2

Add $1$ to both sides of the equation.

$u = 1$

Step 21.7

The final solution is all the values that make $(u - 8) (u - 1) = 0$ true.

$u = 8, 1$

Step 21.8

Substitute $\cos (x)$ for $u$ .

$\cos (x) = 8, 1$

Step 21.9

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

$\cos (x) = 8$

$\cos (x) = 1$

Step 21.10

Solve for $x$ in $\cos (x) = 8$ .

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Step 21.10.1

The range of cosine is $- 1 \leq y \leq 1$ . Since $8$ does not fall in this range, there is no solution.

No solution

Step 21.11

Solve for $x$ in $\cos (x) = 1$ .

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Step 21.11.1

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

$x = \arccos (1)$

Step 21.11.2

Simplify the right side.

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Step 21.11.2.1

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

$x = 0$

Step 21.11.3

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.

$x = 2 π - 0$

Step 21.11.4

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 21.11.5

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

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Step 21.11.5.1

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

$\frac{2 π}{| b |}$

Step 21.11.5.2

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

$\frac{2 π}{| 1 |}$

Step 21.11.5.3

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

$\frac{2 π}{1}$

Step 21.11.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 21.11.6

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 21.12

List all of the solutions.

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

Step 21.13

Consolidate the answers.

$x = 2 π n$ , for any integer $n$