Solve for θ in Degrees 11sin(theta)+2=4sin(theta)+2

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Trigonometry

$11 \sin (θ) + 2 = 4 \sin (θ) + 2$

Move all terms containing $\sin (θ)$ to the left side of the equation.

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Subtract $4 \sin (θ)$ from both sides of the equation.

$11 \sin (θ) + 2 - 4 \sin (θ) = 2$

Subtract $4 \sin (θ)$ from $11 \sin (θ)$ .

$7 \sin (θ) + 2 = 2$

Move all terms not containing $\sin (θ)$ to the right side of the equation.

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Subtract $2$ from both sides of the equation.

$7 \sin (θ) = 2 - 2$

Subtract $2$ from $2$ .

$7 \sin (θ) = 0$

Divide each term in $7 \sin (θ) = 0$ by $7$ and simplify.

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Divide each term in $7 \sin (θ) = 0$ by $7$ .

$\frac{7 \sin (θ)}{7} = \frac{0}{7}$

Simplify the left side.

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Cancel the common factor of $7$ .

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Cancel the common factor.

$\frac{7 \sin (θ)}{7} = \frac{0}{7}$

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{0}{7}$

Simplify the right side.

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Divide $0$ by $7$ .

$\sin (θ) = 0$

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 $n$

Consolidate the answers.

$θ = 180 n$ , for any integer $n$

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