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

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

Rewrite the equation as $- 2 \sin (2 x) = \sin (4 x)$ .

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

Divide each term by $- 2$ and simplify.

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

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

Cancel the common factor of $- 2$ .

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Cancel the common factor.

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

Divide $\sin (2 x)$ by $1$ .

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

Move the negative in front of the fraction.

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

Move all terms containing $x$ to the left side of the equation.

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Add $- \frac{\sin (4 x)}{2}$ to both sides of the equation.

$\sin (2 x) + \frac{\sin (4 x)}{2} = 0$

Simplify the left side of the equation.

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Simplify each term.

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Apply the sine double–angle identity.

$2 \sin (x) \cos (x) + \frac{\sin (4 x)}{2} = 0$

Simplify the numerator.

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Factor $2$ out of $4 x$ .

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

Apply the sine double–angle identity.

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

Use the double–angle identity to transform $\cos (2 x)$ to $\cos^{2} (x) - \sin^{2} (x)$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \cos (x)$ and $b = \sin (x)$ .

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

Multiply $2$ by $2$ .

$2 \sin (x) \cos (x) + \frac{4 (\sin (x) \cos (x)) (\cos (x) + \sin (x)) (\cos (x) - \sin (x))}{2} = 0$

Simplify the numerator.

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Rewrite.

$2 \sin (x) \cos (x) + \frac{\sin (4 x)}{2} = 0$

Factor $2$ out of $4 x$ .

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

Apply the sine double–angle identity.

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

Use the double–angle identity to transform $\cos (2 x)$ to $\cos^{2} (x) - \sin^{2} (x)$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \cos (x)$ and $b = \sin (x)$ .

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

Multiply $2$ by $2$ .

$2 \sin (x) \cos (x) + \frac{4 (\sin (x) \cos (x)) (\cos (x) + \sin (x)) (\cos (x) - \sin (x))}{2} = 0$

Remove unnecessary parentheses.

$2 \sin (x) \cos (x) + \frac{4 \sin (x) \cos (x) (\cos (x) + \sin (x)) (\cos (x) - \sin (x))}{2} = 0$

Cancel the common factor of $4$ and $2$ .

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Factor $2$ out of $4 \sin (x) \cos (x) (\cos (x) + \sin (x)) (\cos (x) - \sin (x))$ .

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

Cancel the common factors.

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Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2 \sin (x) \cos (x) (\cos (x) + \sin (x)) (\cos (x) - \sin (x))$ by $1$ .

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

Apply the sine double–angle identity.

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

Apply the distributive property.

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

Expand $(\sin (2 x) \cos (x) + \sin (2 x) \sin (x)) (\cos (x) - \sin (x))$ using the FOIL Method.

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Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Combine the opposite terms in $\sin (2 x) \cos (x) \cos (x) + \sin (2 x) \cos (x) (- \sin (x)) + \sin (2 x) \sin (x) \cos (x) + \sin (2 x) \sin (x) (- \sin (x))$ .

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Reorder the factors in the terms $\sin (2 x) \cos (x) (- \sin (x))$ and $\sin (2 x) \sin (x) \cos (x)$ .

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

Add $- \cos (x) \sin (x) \sin (2 x)$ and $\cos (x) \sin (x) \sin (2 x)$ .

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

Add $\sin (2 x) \cos (x) \cos (x)$ and $0$ .

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

Simplify each term.

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Multiply $\sin (2 x) \cos (x) \cos (x)$ .

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

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

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

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

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

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

Add $1$ and $1$ .

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

Rewrite using the commutative property of multiplication.

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

Multiply $- \sin (2 x) \sin (x) \sin (x)$ .

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

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

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

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

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

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

Add $1$ and $1$ .

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

Apply the sine double–angle identity.

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

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

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

Multiply by $1$ .

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

Simplify with factoring out.

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

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

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

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

Apply pythagorean identity.

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

Add $\sin (2 x) \cos^{2} (x)$ and $\sin (2 x) \cos^{2} (x)$ .

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

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

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

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

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

Divide $0$ by $2$ .

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

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

$\sin (2 x) = 0$

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

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

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Set the first factor equal to $0$ .

$\sin (2 x) = 0$

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

$2 x = \arcsin (0)$

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

$2 x = 0$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Divide $0$ by $2$ .

$x = 0$

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.

$2 x = π - 0$

Simplify the expression to find the second solution.

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Multiply $- 1$ by $0$ .

$2 x = π + 0$

Add $π$ and $0$ .

$2 x = π$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = π$ by $2$ .

$\frac{2 x}{2} = \frac{π}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Divide $x$ by $1$ .

$x = \frac{π}{2}$

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 $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

Solve the equation.

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

$\frac{2 π}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 π}{2}$

Divide $π$ by $1$ .

$π$

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

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

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

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Set the next factor equal to $0$ .

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

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

$\cos (x) = \pm \sqrt{0}$

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 $0$ as $0^{2}$ .

$\cos (x) = \pm \sqrt{0^{2}}$

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

$\cos (x) = \pm 0$

$\pm 0$ is 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 $n$

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

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

Consolidate the answers.

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

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

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

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

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

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

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

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