Find All Complex Solutions sin(2theta)+sin(4theta)=0

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Trigonometry

$\sin (2 θ) + \sin (4 θ) = 0$

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

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Factor $2 \sin (θ) \cos (θ)$ out of $2 \sin (θ) \cos (θ)$ .

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

Factor $2 \sin (θ) \cos (θ)$ out of $4 \sin (θ) \cos^{3} (θ)$ .

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

Factor $2 \sin (θ) \cos (θ)$ out of $- 4 \sin^{3} (θ) \cos (θ)$ .

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

Factor $2 \sin (θ) \cos (θ)$ out of $2 \sin (θ) \cos (θ) \cdot 1 + 2 \sin (θ) \cos (θ) (2 \cos^{2} (θ))$ .

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

Factor $2 \sin (θ) \cos (θ)$ out of $2 \sin (θ) \cos (θ) (1 + 2 \cos^{2} (θ)) + 2 \sin (θ) \cos (θ) (- 2 \sin^{2} (θ))$ .

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

Divide each term in $2 \sin (θ) \cos (θ) (1 + 2 \cos^{2} (θ) - 2 \sin^{2} (θ)) = 0$ by $2$ .

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

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 \sin (θ) \cos (θ) (1 + 2 \cos^{2} (θ) - 2 \sin^{2} (θ))}{2}$

Divide $\sin (θ) \cos (θ) (1 + 2 \cos^{2} (θ) - 2 \sin^{2} (θ))$ by $1$ .

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

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

Divide $0$ by $2$ .

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

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

$\sin (θ) = 0$

$\cos (θ) = 0$

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

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$\sin (θ) = 0$

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

$θ = \arcsin (0)$

The exact value of $\arcsin (0)$ is $0$ .

$θ = 0$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$θ = π - 0$

Subtract $0$ from $π$ .

$θ = π$

Find the period.

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

Solve the equation.

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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

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

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$\cos (θ) = 0$

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

$θ = \arccos (0)$

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$θ = \frac{π}{2}$

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.

$θ = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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

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

Combine fractions.

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Combine $2 π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

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Multiply $2$ by $2$ .

$θ = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$θ = \frac{3 π}{2}$

Find the period.

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

Solve the equation.

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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

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

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

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

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

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

Simplify each term.

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Apply the distributive property.

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

Multiply $2$ by $1$ .

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

Multiply $- 1$ by $2$ .

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

Simplify by adding terms.

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Add $1$ and $2$ .

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

Subtract $2 \sin^{2} (θ)$ from $- 2 \sin^{2} (θ)$ .

$3 - 4 \sin^{2} (θ) = 0$

Reorder the polynomial.

$- 4 \sin^{2} (θ) + 3 = 0$

Subtract $3$ from both sides of the equation.

$- 4 \sin^{2} (θ) = - 3$

Divide each term by $- 4$ and simplify.

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Divide each term in $- 4 \sin^{2} (θ) = - 3$ by $- 4$ .

$\frac{- 4 \sin^{2} (θ)}{- 4} = \frac{- 3}{- 4}$

Cancel the common factor of $- 4$ .

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Cancel the common factor.

$\frac{- 4 \sin^{2} (θ)}{- 4} = \frac{- 3}{- 4}$

Divide $\sin^{2} (θ)$ by $1$ .

$\sin^{2} (θ) = \frac{- 3}{- 4}$

Dividing two negative values results in a positive value.

$\sin^{2} (θ) = \frac{3}{4}$

Take the $square$ root of both sides of the $equation$ to eliminate the exponent on the left side.

$\sin (θ) = \pm \sqrt{\frac{3}{4}}$

The complete solution is the result of both the positive and negative portions of the solution.

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Simplify the right side of the equation.

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Rewrite $\sqrt{\frac{3}{4}}$ as $\frac{\sqrt{3}}{\sqrt{4}}$ .

$\sin (θ) = \pm \frac{\sqrt{3}}{\sqrt{4}}$

Simplify the denominator.

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

$\sin (θ) = \pm \frac{\sqrt{3}}{\sqrt{2^{2}}}$

Pull terms out from under the radical.

$\sin (θ) = \pm \frac{\sqrt{3}}{| 2 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\sin (θ) = \pm \frac{\sqrt{3}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$\sin (θ) = \frac{\sqrt{3}}{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$\sin (θ) = - \frac{\sqrt{3}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$\sin (θ) = \frac{\sqrt{3}}{2}, - \frac{\sqrt{3}}{2}$

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

$\sin (θ) = \frac{\sqrt{3}}{2}$

$\sin (θ) = - \frac{\sqrt{3}}{2}$

Set up the equation to solve for $θ$ .

$\sin (θ) = \frac{\sqrt{3}}{2}$

Solve the equation for $θ$ .

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

$θ = \arcsin (\frac{\sqrt{3}}{2})$

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

$θ = \frac{π}{3}$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$θ = π - \frac{π}{3}$

Simplify $π - \frac{π}{3}$ .

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

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

Combine fractions.

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Combine $π$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 3 - π}{3}$

Simplify the numerator.

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Move $3$ to the left of $π$ .

$θ = \frac{3 \cdot π - π}{3}$

Subtract $π$ from $3 π$ .

$θ = \frac{2 π}{3}$

Find the period.

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

Solve the equation.

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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

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

Set up the equation to solve for $θ$ .

$\sin (θ) = - \frac{\sqrt{3}}{2}$

Solve the equation for $θ$ .

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

$θ = \arcsin (- \frac{\sqrt{3}}{2})$

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

$θ = - \frac{π}{3}$

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$θ = 2 π + \frac{π}{3} + π$

Simplify the expression to find the second solution.

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Simplify $2 π + \frac{π}{3} + π$ .

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

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

Combine fractions.

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Combine $2 π$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 + π}{3} + π$

Simplify the numerator.

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Multiply $3$ by $2$ .

$θ = \frac{6 π + π}{3} + π$

Add $6 π$ and $π$ .

$θ = \frac{7 π}{3} + π$

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

$θ = \frac{7 π}{3} + π \cdot \frac{3}{3}$

Combine fractions.

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Combine $π$ and $\frac{3}{3}$ .

$θ = \frac{7 π}{3} + \frac{π \cdot 3}{3}$

Combine the numerators over the common denominator.

$θ = \frac{7 π + π \cdot 3}{3}$

Simplify the numerator.

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Move $3$ to the left of $π$ .

$θ = \frac{7 π + 3 \cdot π}{3}$

Add $7 π$ and $3 π$ .

$θ = \frac{10 π}{3}$

Subtract $2 π$ from $\frac{10 π}{3}$ .

$θ = \frac{10 π}{3} - 2 π$

The resulting angle of $\frac{4 π}{3}$ is positive, less than $2 π$ , and coterminal with $\frac{10 π}{3}$ .

$θ = \frac{4 π}{3}$

Find the period.

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

Solve the equation.

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

Add $2 π$ to every negative angle to get positive angles.

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Add $2 π$ to $- \frac{π}{3}$ to find the positive angle.

$- \frac{π}{3} + 2 π$

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

$2 π \cdot \frac{3}{3} - \frac{π}{3}$

Combine fractions.

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Combine $2 π$ and $\frac{3}{3}$ .

$\frac{2 π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 3 - π}{3}$

Simplify the numerator.

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Multiply $3$ by $2$ .

$\frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$\frac{5 π}{3}$

List the new angles.

$θ = \frac{5 π}{3}$

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

$θ = \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

List all of the results found in the previous steps.

$θ = \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

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

$θ = 2 π n, π + 2 π n, \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n, \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Consolidate the answers.

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

$θ = π n, \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n, \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Consolidate $\frac{π}{2} + 2 π n$ and $\frac{3 π}{2} + 2 π n$ to $\frac{π}{2} + π n$ .

$θ = π n, \frac{π}{2} + π n, \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Consolidate $π n$ and $\frac{π}{2} + π n$ to $\frac{π n}{2}$ .

$θ = \frac{π n}{2}, \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Consolidate $\frac{π}{3} + 2 π n$ and $\frac{4 π}{3} + 2 π n$ to $\frac{π}{3} + π n$ .

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

Consolidate $\frac{2 π}{3} + 2 π n$ and $\frac{5 π}{3} + 2 π n$ to $\frac{2 π}{3} + π n$ .

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

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