Solve for C in Degrees 10cos(C)+1=cos(C)-1

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Trigonometry

$10 \cos (C) + 1 = \cos (C) - 1$

Move all terms containing $\cos (C)$ to the left side of the equation.

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Subtract $\cos (C)$ from both sides of the equation.

$10 \cos (C) + 1 - \cos (C) = - 1$

Subtract $\cos (C)$ from $10 \cos (C)$ .

$9 \cos (C) + 1 = - 1$

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

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Subtract $1$ from both sides of the equation.

$9 \cos (C) = - 1 - 1$

Subtract $1$ from $- 1$ .

$9 \cos (C) = - 2$

Divide each term in $9 \cos (C) = - 2$ by $9$ and simplify.

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Divide each term in $9 \cos (C) = - 2$ by $9$ .

$\frac{9 \cos (C)}{9} = \frac{- 2}{9}$

Simplify the left side.

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Cancel the common factor of $9$ .

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Cancel the common factor.

$\frac{9 \cos (C)}{9} = \frac{- 2}{9}$

Divide $\cos (C)$ by $1$ .

$\cos (C) = \frac{- 2}{9}$

Simplify the right side.

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Move the negative in front of the fraction.

$\cos (C) = - \frac{2}{9}$

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

$C = \arccos (- \frac{2}{9})$

Simplify the right side.

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Evaluate $\arccos (- \frac{2}{9})$ .

$C = 102.8395884$

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

$C = 360 - 102.8395884$

Subtract $102.8395884$ from $360$ .

$C = 257.16041159$

Find the period of $\cos (C)$ .

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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 $\cos (C)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$C = 102.8395884 + 360 n, 257.16041159 + 360 n$ , for any integer $n$

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