Solve for x sin(2x)cos(x)+cos(2x)sin(x)=0

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

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

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

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

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

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

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

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

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

$\sin (x) = 0$

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

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 $0$ .

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

Replace the $3 \cos^{2} (x)$ with $3 (1 - \sin^{2} (x))$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

$3 (1 - \sin^{2} (x)) - \sin^{2} (x) = 0$

Simplify each term.

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Apply the distributive property.

$3 \cdot 1 + 3 (- \sin^{2} (x)) - \sin^{2} (x) = 0$

Multiply $3$ by $1$ .

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

Multiply $- 1$ by $3$ .

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

Subtract $\sin^{2} (x)$ from $- 3 \sin^{2} (x)$ .

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

Reorder the polynomial.

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

Subtract $3$ from both sides of the equation.

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

Divide each term by $- 4$ and simplify.

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

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

Cancel the common factor of $- 4$ .

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Cancel the common factor.

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

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

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

Dividing two negative values results in a positive value.

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

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

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

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

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

Simplify the denominator.

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

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

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

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

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.

$\sin (x) = \frac{\sqrt{3}}{2}$

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

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

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

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

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

$\sin (x) = \frac{\sqrt{3}}{2}$

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

Set up the equation to solve for $x$ .

$\sin (x) = \frac{\sqrt{3}}{2}$

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.

$x = \arcsin (\frac{\sqrt{3}}{2})$

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

$x = \frac{π}{3}$

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.

$x = π - \frac{π}{3}$

Simplify $π - \frac{π}{3}$ .

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To write $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = \frac{π}{1} \cdot \frac{3}{3} - \frac{π}{3}$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{π \cdot 3}{1 \cdot 3} - \frac{π}{3}$

Multiply $3$ by $1$ .

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

Combine the numerators over the common denominator.

$x = \frac{π \cdot 3 - π}{3}$

Simplify the numerator.

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Move $3$ to the left of $π$ .

$x = \frac{3 \cdot π - π}{3}$

Subtract $π$ from $3 π$ .

$x = \frac{2 π}{3}$

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

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

Set up the equation to solve for $x$ .

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

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.

$x = \arcsin (- \frac{\sqrt{3}}{2})$

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

$x = - \frac{π}{3}$

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.

$x = 2 π + \frac{π}{3} + π$

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 $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = \frac{2 π}{1} \cdot \frac{3}{3} + \frac{π}{3} + π$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{2 π \cdot 3}{1 \cdot 3} + \frac{π}{3} + π$

Multiply $3$ by $1$ .

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

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 + π}{3} + π$

Simplify the numerator.

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Multiply $3$ by $2$ .

$x = \frac{6 π + π}{3} + π$

Add $6 π$ and $π$ .

$x = \frac{7 π}{3} + π$

To write $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = \frac{7 π}{3} + \frac{π}{1} \cdot \frac{3}{3}$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{7 π}{3} + \frac{π \cdot 3}{1 \cdot 3}$

Multiply $3$ by $1$ .

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

Combine the numerators over the common denominator.

$x = \frac{7 π + π \cdot 3}{3}$

Simplify the numerator.

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Move $3$ to the left of $π$ .

$x = \frac{7 π + 3 \cdot π}{3}$

Add $7 π$ and $3 π$ .

$x = \frac{10 π}{3}$

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

$x = \frac{10 π}{3} - 2 π$

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

$x = \frac{4 π}{3}$

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 $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{2 π}{1} \cdot \frac{3}{3} - \frac{π}{3}$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$\frac{2 π \cdot 3}{1 \cdot 3} - \frac{π}{3}$

Multiply $3$ by $1$ .

$\frac{2 π \cdot 3}{3} - \frac{π}{3}$

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.

$x = \frac{5 π}{3}$

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

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

List all of the results found in the previous steps.

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

The final solution is all the values that make $\sin (x) (3 \cos^{2} (x) - \sin^{2} (x)) = 0$ true.

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

Consolidate the answers.

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

Solve for x sin(2x)cos(x)+cos(2x)sin(x)=0

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