Solve for A in Degrees 5cos(A)+8=3cos(A)+6

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Trigonometry

$5 \cos (A) + 8 = 3 \cos (A) + 6$

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

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Subtract $3 \cos (A)$ from both sides of the equation.

$5 \cos (A) + 8 - 3 \cos (A) = 6$

Subtract $3 \cos (A)$ from $5 \cos (A)$ .

$2 \cos (A) + 8 = 6$

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

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Subtract $8$ from both sides of the equation.

$2 \cos (A) = 6 - 8$

Subtract $8$ from $6$ .

$2 \cos (A) = - 2$

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

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

$\frac{2 \cos (A)}{2} = \frac{- 2}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 \cos (A)}{2} = \frac{- 2}{2}$

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

$\cos (A) = \frac{- 2}{2}$

Simplify the right side.

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Divide $- 2$ by $2$ .

$\cos (A) = - 1$

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

$A = \arccos (- 1)$

Simplify the right side.

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

$A = 180$

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.

$A = 360 - 180$

Subtract $180$ from $360$ .

$A = 180$

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

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

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

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