Precalculus Examples

$2 \sin^{2} (θ) = 3 (1 - \cos (- θ))$

Step 1

Simplify the right side.

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Step 1.1

Simplify $3 (1 - \cos (- θ))$ .

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Step 1.1.1

Since $\cos (- θ)$ is an even function, rewrite $\cos (- θ)$ as $\cos (θ)$ .

$2 \sin^{2} (θ) = 3 (1 - \cos (θ))$

Step 1.1.2

Apply the distributive property.

$2 \sin^{2} (θ) = 3 \cdot 1 + 3 (- \cos (θ))$

Step 1.1.3

Multiply.

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Step 1.1.3.1

Multiply $3$ by $1$ .

$2 \sin^{2} (θ) = 3 + 3 (- \cos (θ))$

Step 1.1.3.2

Multiply $- 1$ by $3$ .

$2 \sin^{2} (θ) = 3 - 3 \cos (θ)$

Step 2

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

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Step 2.1

Subtract $3$ from both sides of the equation.

$2 \sin^{2} (θ) - 3 = - 3 \cos (θ)$

Step 2.2

Add $3 \cos (θ)$ to both sides of the equation.

$2 \sin^{2} (θ) - 3 + 3 \cos (θ) = 0$

Step 3

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

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

Step 4

Simplify each term.

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Step 4.1

Apply the distributive property.

$2 \cdot 1 + 2 (- \cos^{2} (θ)) - 3 + 3 \cos (θ) = 0$

Step 4.2

Multiply $2$ by $1$ .

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

Step 4.3

Multiply $- 1$ by $2$ .

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

Step 5

Subtract $3$ from $2$ .

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

Step 6

Reorder the polynomial.

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

Step 7

Substitute $u$ for $\cos (θ)$ .

$- 2 {(u)}^{2} + 3 (u) - 1 = 0$

Step 8

Factor the left side of the equation.

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Step 8.1

Factor $- 1$ out of $- 2 u^{2} + 3 u - 1$ .

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Step 8.1.1

Factor $- 1$ out of $- 2 u^{2}$ .

$- (2 u^{2}) + 3 u - 1 = 0$

Step 8.1.2

Factor $- 1$ out of $3 u$ .

$- (2 u^{2}) - (- 3 u) - 1 = 0$

Step 8.1.3

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

$- (2 u^{2}) - (- 3 u) - 1 \cdot 1 = 0$

Step 8.1.4

Factor $- 1$ out of $- (2 u^{2}) - (- 3 u)$ .

$- (2 u^{2} - 3 u) - 1 \cdot 1 = 0$

Step 8.1.5

Factor $- 1$ out of $- (2 u^{2} - 3 u) - 1 (1)$ .

$- (2 u^{2} - 3 u + 1) = 0$

Step 8.2

Factor.

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Step 8.2.1

Factor by grouping.

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Step 8.2.1.1

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 $b = - 3$ .

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Step 8.2.1.1.1

Factor $- 3$ out of $- 3 u$ .

$- (2 u^{2} - 3 u + 1) = 0$

Step 8.2.1.1.2

Rewrite $- 3$ as $- 1$ plus $- 2$

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

Step 8.2.1.1.3

Apply the distributive property.

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

Step 8.2.1.2

Factor out the greatest common factor from each group.

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Step 8.2.1.2.1

Group the first two terms and the last two terms.

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

Step 8.2.1.2.2

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

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

Step 8.2.1.3

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

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

Step 8.2.2

Remove unnecessary parentheses.

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

Step 9

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 10

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

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Step 10.1

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

$2 u - 1 = 0$

Step 10.2

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

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Step 10.2.1

Add $1$ to both sides of the equation.

$2 u = 1$

Step 10.2.2

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

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Step 10.2.2.1

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

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

Step 10.2.2.2

Simplify the left side.

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Step 10.2.2.2.1

Cancel the common factor of $2$ .

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Step 10.2.2.2.1.1

Cancel the common factor.

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

Step 10.2.2.2.1.2

Divide $u$ by $1$ .

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

Step 11

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

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Step 11.1

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

$u - 1 = 0$

Step 11.2

Add $1$ to both sides of the equation.

$u = 1$

Step 12

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

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

Step 13

Substitute $\cos (θ)$ for $u$ .

$\cos (θ) = \frac{1}{2}, 1$

Step 14

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

$\cos (θ) = \frac{1}{2}$

$\cos (θ) = 1$

Step 15

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

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Step 15.1

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

$θ = \arccos (\frac{1}{2})$

Step 15.2

Simplify the right side.

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Step 15.2.1

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

$θ = \frac{π}{3}$

Step 15.3

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 15.4

Simplify $2 π - \frac{π}{3}$ .

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Step 15.4.1

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 15.4.2

Combine fractions.

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Step 15.4.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 15.4.2.2

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - π}{3}$

Step 15.4.3

Simplify the numerator.

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Step 15.4.3.1

Multiply $3$ by $2$ .

$θ = \frac{6 π - π}{3}$

Step 15.4.3.2

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{3}$

Step 15.5

Find the period of $\cos (θ)$ .

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Step 15.5.1

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

$\frac{2 π}{| b |}$

Step 15.5.2

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

$\frac{2 π}{| 1 |}$

Step 15.5.3

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

$\frac{2 π}{1}$

Step 15.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 15.6

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 16

Solve for $θ$ in $\cos (θ) = 1$ .

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Step 16.1

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

$θ = \arccos (1)$

Step 16.2

Simplify the right side.

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Step 16.2.1

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

$θ = 0$

Step 16.3

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.

$θ = 2 π - 0$

Step 16.4

Subtract $0$ from $2 π$ .

$θ = 2 π$

Step 16.5

Find the period of $\cos (θ)$ .

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Step 16.5.1

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

$\frac{2 π}{| b |}$

Step 16.5.2

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

$\frac{2 π}{| 1 |}$

Step 16.5.3

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

$\frac{2 π}{1}$

Step 16.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 16.6

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

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

Step 17

List all of the solutions.

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

Step 18

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

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