Solve for C in Degrees 6cos(C)-4=cos(C)-9

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Trigonometry

$6 \cos (C) - 4 = \cos (C) - 9$

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.

$6 \cos (C) - 4 - \cos (C) = - 9$

Subtract $\cos (C)$ from $6 \cos (C)$ .

$5 \cos (C) - 4 = - 9$

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

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Add $4$ to both sides of the equation.

$5 \cos (C) = - 9 + 4$

Add $- 9$ and $4$ .

$5 \cos (C) = - 5$

Divide each term in $5 \cos (C) = - 5$ by $5$ and simplify.

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Divide each term in $5 \cos (C) = - 5$ by $5$ .

$\frac{5 \cos (C)}{5} = \frac{- 5}{5}$

Simplify the left side.

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Cancel the common factor of $5$ .

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Cancel the common factor.

$\frac{5 \cos (C)}{5} = \frac{- 5}{5}$

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

$\cos (C) = \frac{- 5}{5}$

Simplify the right side.

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Divide $- 5$ by $5$ .

$\cos (C) = - 1$

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

$C = \arccos (- 1)$

Simplify the right side.

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The exact value of $\arccos (- 1)$ is $180$ .

$C = 180$

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.

$C = 360 - 180$

Subtract $180$ from $360$ .

$C = 180$

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.

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

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