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

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Trigonometry

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

Expand the trigonometric functions.

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Simplify each term.

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Apply the sine double-angle identity.

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

Apply the distributive property.

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

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

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Multiply $- 1$ by $- 1$ .

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

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

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

Apply the distributive property.

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

Multiply $\cos^{2} (x)$ by $\cos (x)$ by adding the exponents.

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Move $\cos (x)$ .

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

Multiply $\cos (x)$ by $\cos^{2} (x)$ .

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

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

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

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

Add $1$ and $2$ .

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

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

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

Move $\cos (x)$ to the left side of the equation by adding it to both sides.

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

Simplify the left side.

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Simplify with factoring out.

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

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

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

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

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

Multiply by $1$ .

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

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

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

Factor $\cos (x)$ out of $\cos (x) (3 \sin^{2} (x) - \cos^{2} (x)) + \cos (x) \cdot 1$ .

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

Reorder $- \cos^{2} (x)$ and $1$ .

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

Apply pythagorean identity.

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

Simplify by adding terms.

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Add $3 \sin^{2} (x)$ and $\sin^{2} (x)$ .

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

Rewrite using the commutative property of multiplication.

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

Divide each term in $4 \cos (x) \sin^{2} (x) = 0$ by $4$ .

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

Cancel the common factor of $4$ .

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Cancel the common factor.

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

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

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

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

Divide $0$ by $4$ .

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

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

$\cos (x) = 0$

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

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

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Set the first 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 $0$ .

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

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

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

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 $0^{2}$ .

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

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

$\sin (x) = \pm 0$

$\pm 0$ is 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 $n$

The final solution is all the values that make $\frac{4 \cos (x) \sin^{2} (x)}{4} = \frac{0}{4}$ true.

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

Consolidate the answers.

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

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