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

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Trigonometry

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

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

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

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

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

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

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

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

Rewrite $- 1 \cos (x)$ as $- \cos (x)$ .

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

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

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

$\sin (x) (1 - \cos (x))$

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

Divide $0$ by $2$ .

$\sin (x) (1 - \cos (x)) = 0$

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

$\sin (x) = 0$

$1 - \cos (x) = 0$

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 $0$ .

$1 - \cos (x) = 0$

Subtract $1$ from both sides of the equation.

$- \cos (x) = - 1$

Multiply each term in $- \cos (x) = - 1$ by $- 1$

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Multiply each term in $- \cos (x) = - 1$ by $- 1$ .

$(- \cos (x)) \cdot - 1 = (- 1) \cdot - 1$

Multiply $- \cos (x) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 \cos (x) = (- 1) \cdot - 1$

Multiply $\cos (x)$ by $1$ .

$\cos (x) = (- 1) \cdot - 1$

Multiply $- 1$ by $- 1$ .

$\cos (x) = 1$

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

$x = \arccos (1)$

The exact value of $\arccos (1)$ is $0$ .

$x = 0$

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.

$x = 2 π - 0$

Subtract $0$ from $2 π$ .

$x = 2 π$

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

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

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

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

Consolidate the answers.

$x = π n$ , for any integer $n$

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