Solve for x sin(4x)-sin(2x)=0

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Trigonometry

$\sin (4 x) - \sin (2 x) = 0$

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

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

$2 \sin (x) \cos (x) (2 \cos^{2} (x)) - 4 \sin^{3} (x) \cos (x) - 2 \sin (x) \cos (x) = 0$

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

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

Factor $2 \sin (x) \cos (x)$ out of $- 2 \sin (x) \cos (x)$ .

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

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

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

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

Divide $0$ by $2$ .

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

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

$\sin (x) = 0$

$\cos (x) = 0$

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

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

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Set the first factor equal to $0$ .

$\sin (x) = 0$

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

$x = \arcsin (0)$

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

$x = 0$

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.

$x = π - 0$

Subtract $0$ from $π$ .

$x = π$

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 $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

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

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

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Set the next factor equal to $0$ .

$\cos (x) = 0$

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

$x = \arccos (0)$

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

$x = \frac{π}{2}$

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.

$x = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{2 π}{1} \cdot \frac{2}{2} - \frac{π}{2}$

Write each expression with a common denominator of $2$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{2 π \cdot 2}{1 \cdot 2} - \frac{π}{2}$

Multiply $2$ by $1$ .

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

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

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Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

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

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

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

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Set the next factor equal to $0$ .

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

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

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

Simplify each term.

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Apply the distributive property.

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

Multiply $2$ by $1$ .

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

Multiply $- 1$ by $2$ .

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

Simplify by adding terms.

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Subtract $1$ from $2$ .

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

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

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

Subtract $1$ from both sides of the equation.

$- 4 \sin^{2} (x) = - 1$

Divide each term by $- 4$ and simplify.

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Divide each term in $- 4 \sin^{2} (x) = - 1$ by $- 4$ .

$\frac{- 4 \sin^{2} (x)}{- 4} = \frac{- 1}{- 4}$

Cancel the common factor of $- 4$ .

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Cancel the common factor.

$\frac{- 4 \sin^{2} (x)}{- 4} = \frac{- 1}{- 4}$

Divide $\sin^{2} (x)$ by $1$ .

$\sin^{2} (x) = \frac{- 1}{- 4}$

Dividing two negative values results in a positive value.

$\sin^{2} (x) = \frac{1}{4}$

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

$\sin (x) = \pm \sqrt{\frac{1}{4}}$

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 $\sqrt{\frac{1}{4}}$ as $\frac{\sqrt{1}}{\sqrt{4}}$ .

$\sin (x) = \pm \frac{\sqrt{1}}{\sqrt{4}}$

Any root of $1$ is $1$ .

$\sin (x) = \pm \frac{1}{\sqrt{4}}$

Simplify the denominator.

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

$\sin (x) = \pm \frac{1}{\sqrt{2^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$\sin (x) = \pm \frac{1}{2}$

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.

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

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

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

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

$\sin (x) = \frac{1}{2}, - \frac{1}{2}$

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

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

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

Set up the equation to solve for $x$ .

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

Solve the equation for $x$ .

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

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

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

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.

$x = π - \frac{π}{6}$

Simplify $π - \frac{π}{6}$ .

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To write $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = \frac{π}{1} \cdot \frac{6}{6} - \frac{π}{6}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{π \cdot 6}{1 \cdot 6} - \frac{π}{6}$

Multiply $6$ by $1$ .

$x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Simplify the numerator.

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Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

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 $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Set up the equation to solve for $x$ .

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

Solve the equation for $x$ .

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

$x = \arcsin (- \frac{1}{2})$

The exact value of $\arcsin (- \frac{1}{2})$ is $- \frac{π}{6}$ .

$x = - \frac{π}{6}$

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.

$x = 2 π + \frac{π}{6} + π$

Simplify the expression to find the second solution.

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Simplify $2 π + \frac{π}{6} + π$ .

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To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = \frac{2 π}{1} \cdot \frac{6}{6} + \frac{π}{6} + π$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{2 π \cdot 6}{1 \cdot 6} + \frac{π}{6} + π$

Multiply $6$ by $1$ .

$x = \frac{2 π \cdot 6}{6} + \frac{π}{6} + π$

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 6 + π}{6} + π$

Simplify the numerator.

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Multiply $6$ by $2$ .

$x = \frac{12 π + π}{6} + π$

Add $12 π$ and $π$ .

$x = \frac{13 π}{6} + π$

To write $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = \frac{13 π}{6} + \frac{π}{1} \cdot \frac{6}{6}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{13 π}{6} + \frac{π \cdot 6}{1 \cdot 6}$

Multiply $6$ by $1$ .

$x = \frac{13 π}{6} + \frac{π \cdot 6}{6}$

Combine the numerators over the common denominator.

$x = \frac{13 π + π \cdot 6}{6}$

Simplify the numerator.

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Move $6$ to the left of $π$ .

$x = \frac{13 π + 6 \cdot π}{6}$

Add $13 π$ and $6 π$ .

$x = \frac{19 π}{6}$

Subtract $2 π$ from $\frac{19 π}{6}$ .

$x = \frac{19 π}{6} - 2 π$

The resulting angle of $\frac{7 π}{6}$ is positive, less than $2 π$ , and coterminal with $\frac{19 π}{6}$ .

$x = \frac{7 π}{6}$

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 $- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + 2 π$

To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$\frac{2 π}{1} \cdot \frac{6}{6} - \frac{π}{6}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$\frac{2 π \cdot 6}{1 \cdot 6} - \frac{π}{6}$

Multiply $6$ by $1$ .

$\frac{2 π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{6}$

Simplify the numerator.

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Multiply $6$ by $2$ .

$\frac{12 π - π}{6}$

Subtract $π$ from $12 π$ .

$\frac{11 π}{6}$

List the new angles.

$x = \frac{11 π}{6}$

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

$x = \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

List all of the results found in the previous steps.

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

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

$x = 2 π n, π + 2 π n, \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n, \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Consolidate the answers.

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

$x = π n, \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n, \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

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

$x = π n, \frac{π}{2} + π n, \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

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

$x = \frac{π n}{2}, \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Consolidate $\frac{π}{6} + 2 π n$ and $\frac{7 π}{6} + 2 π n$ to $\frac{π}{6} + π n$ .

$x = \frac{π n}{2}, \frac{π}{6} + π n, \frac{5 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Consolidate $\frac{5 π}{6} + 2 π n$ and $\frac{11 π}{6} + 2 π n$ to $\frac{5 π}{6} + π n$ .

$x = \frac{π n}{2}, \frac{π}{6} + π n, \frac{5 π}{6} + π n$ , for any integer $n$

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