Trigonometry Examples

14 \cos (θ) - 5 = 5 \cos (θ) - 5

$14 \cos (θ) - 5 = 5 \cos (θ) - 5$

Step 1

Move all terms containing

\cos (θ)

$\cos (θ)$ to the left side of the equation.

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Step 1.1

Subtract

5 \cos (θ)

$5 \cos (θ)$ from both sides of the equation.

14 \cos (θ) - 5 - 5 \cos (θ) = - 5

$14 \cos (θ) - 5 - 5 \cos (θ) = - 5$

Step 1.2

Subtract

5 \cos (θ)

$5 \cos (θ)$ from

14 \cos (θ)

$14 \cos (θ)$ .

9 \cos (θ) - 5 = - 5

$9 \cos (θ) - 5 = - 5$

9 \cos (θ) - 5 = - 5

$9 \cos (θ) - 5 = - 5$

Step 2

Move all terms not containing

\cos (θ)

$\cos (θ)$ to the right side of the equation.

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Step 2.1

Add

5

$5$ to both sides of the equation.

9 \cos (θ) = - 5 + 5

$9 \cos (θ) = - 5 + 5$

Step 2.2

Add

- 5

$- 5$ and

5

$5$ .

9 \cos (θ) = 0

$9 \cos (θ) = 0$

9 \cos (θ) = 0

$9 \cos (θ) = 0$

Step 3

Divide each term in

9 \cos (θ) = 0

$9 \cos (θ) = 0$ by

9

$9$ and simplify.

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Step 3.1

Divide each term in

9 \cos (θ) = 0

$9 \cos (θ) = 0$ by

9

$9$ .

\frac{9 \cos (θ)}{9} = \frac{0}{9}

$\frac{9 \cos (θ)}{9} = \frac{0}{9}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of

9

$9$ .

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Step 3.2.1.1

Cancel the common factor.

\frac{9 \cos (θ)}{9} = \frac{0}{9}

$\frac{9 \cos (θ)}{9} = \frac{0}{9}$

Step 3.2.1.2

Divide

\cos (θ)

$\cos (θ)$ by

1

$1$ .

\cos (θ) = \frac{0}{9}

$\cos (θ) = \frac{0}{9}$

\cos (θ) = \frac{0}{9}

$\cos (θ) = \frac{0}{9}$

\cos (θ) = \frac{0}{9}

$\cos (θ) = \frac{0}{9}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Divide

0

$0$ by

9

$9$ .

\cos (θ) = 0

$\cos (θ) = 0$

\cos (θ) = 0

$\cos (θ) = 0$

\cos (θ) = 0

$\cos (θ) = 0$

Step 4

Take the inverse cosine of both sides of the equation to extract

θ

$θ$ from inside the cosine.

θ = \arccos (0)

$θ = \arccos (0)$

Step 5

Simplify the right side.

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Step 5.1

The exact value of

\arccos (0)

$\arccos (0)$ is

90

$90$ .

θ = 90

$θ = 90$

θ = 90

$θ = 90$

Step 6

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

360

$360$ to find the solution in the fourth quadrant.

θ = 360 - 90

$θ = 360 - 90$

Step 7

Subtract

90

$90$ from

360

$360$ .

θ = 270

$θ = 270$

Step 8

Find the period of

\cos (θ)

$\cos (θ)$ .

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Step 8.1

The period of the function can be calculated using

\frac{360}{| b |}

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

\frac{360}{| b |}

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

Step 8.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{360}{| 1 |}

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

Step 8.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{360}{1}

$\frac{360}{1}$

Step 8.4

Divide

360

$360$ by

1

$1$ .

360

$360$

360

$360$

Step 9

The period of the

\cos (θ)

$\cos (θ)$ function is

360

$360$ so values will repeat every

360

$360$ degrees in both directions.

θ = 90 + 360 n, 270 + 360 n

$θ = 90 + 360 n, 270 + 360 n$ , for any integer

n

$n$

Step 10

Consolidate the answers.

θ = 90 + 180 n

$θ = 90 + 180 n$ , for any integer

n

$n$