Solve for θ in Degrees sin(theta)-sin(2theta)=0

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Trigonometry

$\sin (θ) - \sin (2 θ) = 0$

Simplify each term.

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Apply the sine double-angle identity.

$\sin (θ) - (2 \sin (θ) \cos (θ)) = 0$

Multiply $2$ by $- 1$ .

$\sin (θ) - 2 \sin (θ) \cos (θ) = 0$

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

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Factor $\sin (θ)$ out of $\sin^{1} (θ)$ .

$\sin (θ) \cdot 1 - 2 \sin (θ) \cos (θ) = 0$

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

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

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

$\sin (θ) (1 - 2 \cos (θ)) = 0$

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

$\sin (θ) = 0$

$1 - 2 \cos (θ) = 0$

Set $\sin (θ)$ equal to $0$ and solve for $θ$ .

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Set $\sin (θ)$ equal to $0$ .

$\sin (θ) = 0$

Solve $\sin (θ) = 0$ for $θ$ .

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

$θ = \arcsin (0)$

Simplify the right side.

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

$θ = 0$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the second quadrant.

$θ = 180 - 0$

Subtract $0$ from $180$ .

$θ = 180$

Find the period of $\sin (θ)$ .

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{360}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

The period of the $\sin (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 360 n, 180 + 360 n$ , for any integer $n$

Set $1 - 2 \cos (θ)$ equal to $0$ and solve for $θ$ .

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Set $1 - 2 \cos (θ)$ equal to $0$ .

$1 - 2 \cos (θ) = 0$

Solve $1 - 2 \cos (θ) = 0$ for $θ$ .

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Subtract $1$ from both sides of the equation.

$- 2 \cos (θ) = - 1$

Divide each term in $- 2 \cos (θ) = - 1$ by $- 2$ and simplify.

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Divide each term in $- 2 \cos (θ) = - 1$ by $- 2$ .

$\frac{- 2 \cos (θ)}{- 2} = \frac{- 1}{- 2}$

Simplify the left side.

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Cancel the common factor of $- 2$ .

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Cancel the common factor.

$\frac{- 2 \cos (θ)}{- 2} = \frac{- 1}{- 2}$

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{- 1}{- 2}$

Simplify the right side.

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Dividing two negative values results in a positive value.

$\cos (θ) = \frac{1}{2}$

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

$θ = \arccos (\frac{1}{2})$

Simplify the right side.

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The exact value of $\arccos (\frac{1}{2})$ is $60$ .

$θ = 60$

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the fourth quadrant.

$θ = 360 - 60$

Subtract $60$ from $360$ .

$θ = 300$

Find the period of $\cos (θ)$ .

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{360}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

The period of the $\cos (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 60 + 360 n, 300 + 360 n$ , for any integer $n$

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

$θ = 360 n, 180 + 360 n, 60 + 360 n, 300 + 360 n$ , for any integer $n$

Consolidate $360 n$ and $180 + 360 n$ to $180 n$ .

$θ = 180 n, 60 + 360 n, 300 + 360 n$ , for any integer $n$

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