Algebra Examples

$- 2 \sin^{2} (θ) + \cos (θ) + 1 = 0$

Step 1

Replace the $- 2 \sin^{2} (θ)$ with $- 2 (1 - \cos^{2} (θ))$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

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

Step 2

Simplify each term.

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Step 2.1

Apply the distributive property.

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

Step 2.2

Multiply $- 2$ by $1$ .

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

Step 2.3

Multiply $- 1$ by $- 2$ .

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

Step 3

Add $- 2$ and $1$ .

$2 \cos^{2} (θ) + \cos (θ) - 1 = 0$

Step 4

Substitute $u$ for $\cos (θ)$ .

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

Step 5

Factor by grouping.

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Step 5.1

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 $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = 1$ .

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Step 5.1.1

Multiply by $1$ .

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

Step 5.1.2

Rewrite $1$ as $- 1$ plus $2$

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

Step 5.1.3

Apply the distributive property.

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

Step 5.2

Factor out the greatest common factor from each group.

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Step 5.2.1

Group the first two terms and the last two terms.

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

Step 5.2.2

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

$u (2 u - 1) + 1 (2 u - 1) = 0$

Step 5.3

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

$(2 u - 1) (u + 1) = 0$

Step 6

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

$2 u - 1 = 0$

$u + 1 = 0$

Step 7

Set $2 u - 1$ equal to $0$ and solve for $u$ .

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Step 7.1

Set $2 u - 1$ equal to $0$ .

$2 u - 1 = 0$

Step 7.2

Solve $2 u - 1 = 0$ for $u$ .

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Step 7.2.1

Add $1$ to both sides of the equation.

$2 u = 1$

Step 7.2.2

Divide each term in $2 u = 1$ by $2$ and simplify.

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Step 7.2.2.1

Divide each term in $2 u = 1$ by $2$ .

$\frac{2 u}{2} = \frac{1}{2}$

Step 7.2.2.2

Simplify the left side.

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Step 7.2.2.2.1

Cancel the common factor of $2$ .

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Step 7.2.2.2.1.1

Cancel the common factor.

$\frac{2 u}{2} = \frac{1}{2}$

Step 7.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{1}{2}$

Step 8

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

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Step 8.1

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

$u + 1 = 0$

Step 8.2

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 9

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

$u = \frac{1}{2}, - 1$

Step 10

Substitute $\cos (θ)$ for $u$ .

$\cos (θ) = \frac{1}{2}, - 1$

Step 11

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

$\cos (θ) = \frac{1}{2}$

$\cos (θ) = - 1$

Step 12

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

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Step 12.1

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

$θ = \arccos (\frac{1}{2})$

Step 12.2

Simplify the right side.

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Step 12.2.1

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

$θ = \frac{π}{3}$

Step 12.3

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.

$θ = 2 π - \frac{π}{3}$

Step 12.4

Simplify $2 π - \frac{π}{3}$ .

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Step 12.4.1

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

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

Step 12.4.2

Combine fractions.

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Step 12.4.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 12.4.2.2

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - π}{3}$

Step 12.4.3

Simplify the numerator.

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Step 12.4.3.1

Multiply $3$ by $2$ .

$θ = \frac{6 π - π}{3}$

Step 12.4.3.2

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{3}$

Step 12.5

Find the period of $\cos (θ)$ .

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Step 12.5.1

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

$\frac{2 π}{| b |}$

Step 12.5.2

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

$\frac{2 π}{| 1 |}$

Step 12.5.3

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

$\frac{2 π}{1}$

Step 12.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 12.6

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

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

Step 13

Solve for $θ$ in $\cos (θ) = - 1$ .

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Step 13.1

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

$θ = \arccos (- 1)$

Step 13.2

Simplify the right side.

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Step 13.2.1

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

$θ = π$

Step 13.3

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.

$θ = 2 π - π$

Step 13.4

Subtract $π$ from $2 π$ .

$θ = π$

Step 13.5

Find the period of $\cos (θ)$ .

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Step 13.5.1

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

$\frac{2 π}{| b |}$

Step 13.5.2

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

$\frac{2 π}{| 1 |}$

Step 13.5.3

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

$\frac{2 π}{1}$

Step 13.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 13.6

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

$θ = π + 2 π n$ , for any integer $n$

Step 14

List all of the solutions.

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

Step 15

Consolidate the answers.

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