Trigonometry Examples

6 {tan}^{2} (θ) - 10 tan (θ) + 1 = - 5 tan (θ)

$6 \tan^{2} (θ) - 10 \tan (θ) + 1 = - 5 \tan (θ)$

Step 1

Add

5 tan (θ)

$5 \tan (θ)$ to both sides of the equation.

6 {tan}^{2} (θ) - 10 tan (θ) + 1 + 5 tan (θ) = 0

$6 \tan^{2} (θ) - 10 \tan (θ) + 1 + 5 \tan (θ) = 0$

Step 2

Add

- 10 tan (θ)

$- 10 \tan (θ)$ and

5 tan (θ)

$5 \tan (θ)$ .

6 {tan}^{2} (θ) + 1 - 5 tan (θ) = 0

$6 \tan^{2} (θ) + 1 - 5 \tan (θ) = 0$

Step 3

Factor by grouping.

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Step 3.1

Reorder terms.

6 {tan}^{2} (θ) - 5 tan (θ) + 1 = 0

$6 \tan^{2} (θ) - 5 \tan (θ) + 1 = 0$

Step 3.2

For a polynomial of the form

a x^{2} + b x + c

$a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is

a \cdot c = 6 \cdot 1 = 6

$a \cdot c = 6 \cdot 1 = 6$ and whose sum is

b = - 5

$b = - 5$ .

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Step 3.2.1

Factor

- 5

$- 5$ out of

- 5 tan (θ)

$- 5 \tan (θ)$ .

6 {tan}^{2} (θ) - 5 tan (θ) + 1 = 0

$6 \tan^{2} (θ) - 5 \tan (θ) + 1 = 0$

Step 3.2.2

Rewrite

- 5

$- 5$ as

- 2

$- 2$ plus

- 3

$- 3$

6 {tan}^{2} (θ) + (- 2 - 3) tan (θ) + 1 = 0

$6 \tan^{2} (θ) + (- 2 - 3) \tan (θ) + 1 = 0$

Step 3.2.3

Apply the distributive property.

6 {tan}^{2} (θ) - 2 tan (θ) - 3 tan (θ) + 1 = 0

$6 \tan^{2} (θ) - 2 \tan (θ) - 3 \tan (θ) + 1 = 0$

6 {tan}^{2} (θ) - 2 tan (θ) - 3 tan (θ) + 1 = 0

$6 \tan^{2} (θ) - 2 \tan (θ) - 3 \tan (θ) + 1 = 0$

Step 3.3

Factor out the greatest common factor from each group.

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Step 3.3.1

Group the first two terms and the last two terms.

6 {tan}^{2} (θ) - 2 tan (θ) - 3 tan (θ) + 1 = 0

$6 \tan^{2} (θ) - 2 \tan (θ) - 3 \tan (θ) + 1 = 0$

Step 3.3.2

Factor out the greatest common factor (GCF) from each group.

2 tan (θ) (3 tan (θ) - 1) - (3 tan (θ) - 1) = 0

$2 \tan (θ) (3 \tan (θ) - 1) - (3 \tan (θ) - 1) = 0$

2 tan (θ) (3 tan (θ) - 1) - (3 tan (θ) - 1) = 0

$2 \tan (θ) (3 \tan (θ) - 1) - (3 \tan (θ) - 1) = 0$

Step 3.4

Factor the polynomial by factoring out the greatest common factor,

3 tan (θ) - 1

$3 \tan (θ) - 1$ .

(3 tan (θ) - 1) (2 tan (θ) - 1) = 0

$(3 \tan (θ) - 1) (2 \tan (θ) - 1) = 0$

(3 tan (θ) - 1) (2 tan (θ) - 1) = 0

$(3 \tan (θ) - 1) (2 \tan (θ) - 1) = 0$

Step 4

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

3 tan (θ) - 1 = 0

$3 \tan (θ) - 1 = 0$

2 tan (θ) - 1 = 0

$2 \tan (θ) - 1 = 0$

Step 5

Set

3 tan (θ) - 1

$3 \tan (θ) - 1$ equal to

0

$0$ and solve for

θ

$θ$ .

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Step 5.1

Set

3 tan (θ) - 1

$3 \tan (θ) - 1$ equal to

0

$0$ .

3 tan (θ) - 1 = 0

$3 \tan (θ) - 1 = 0$

Step 5.2

Solve

3 tan (θ) - 1 = 0

$3 \tan (θ) - 1 = 0$ for

θ

$θ$ .

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Step 5.2.1

Add

1

$1$ to both sides of the equation.

3 tan (θ) = 1

$3 \tan (θ) = 1$

Step 5.2.2

Divide each term in

3 tan (θ) = 1

$3 \tan (θ) = 1$ by

3

$3$ and simplify.

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Step 5.2.2.1

Divide each term in

3 tan (θ) = 1

$3 \tan (θ) = 1$ by

3

$3$ .

\frac{3 tan (θ)}{3} = \frac{1}{3}

$\frac{3 \tan (θ)}{3} = \frac{1}{3}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of

3

$3$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

\frac{3 tan (θ)}{3} = \frac{1}{3}

$\frac{3 \tan (θ)}{3} = \frac{1}{3}$

Step 5.2.2.2.1.2

Divide

tan (θ)

$\tan (θ)$ by

1

$1$ .

tan (θ) = \frac{1}{3}

$\tan (θ) = \frac{1}{3}$

tan (θ) = \frac{1}{3}

$\tan (θ) = \frac{1}{3}$

tan (θ) = \frac{1}{3}

$\tan (θ) = \frac{1}{3}$

tan (θ) = \frac{1}{3}

$\tan (θ) = \frac{1}{3}$

Step 5.2.3

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

θ

$θ$ from inside the tangent.

θ = arctan (\frac{1}{3})

$θ = \arctan (\frac{1}{3})$

Step 5.2.4

Simplify the right side.

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Step 5.2.4.1

Evaluate

arctan (\frac{1}{3})

$\arctan (\frac{1}{3})$ .

θ = 18.43494882

$θ = 18.43494882$

θ = 18.43494882

$θ = 18.43494882$

Step 5.2.5

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

180

$180$ to find the solution in the fourth quadrant.

θ = 180 + 18.43494882

$θ = 180 + 18.43494882$

Step 5.2.6

Add

180

$180$ and

18.43494882

$18.43494882$ .

θ = 198.43494882

$θ = 198.43494882$

Step 5.2.7

Find the period of

tan (θ)

$\tan (θ)$ .

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Step 5.2.7.1

The period of the function can be calculated using

\frac{180}{| b |}

$\frac{180}{| b |}$ .

\frac{180}{| b |}

$\frac{180}{| b |}$

Step 5.2.7.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{180}{| 1 |}

$\frac{180}{| 1 |}$

Step 5.2.7.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{180}{1}

$\frac{180}{1}$

Step 5.2.7.4

Divide

180

$180$ by

1

$1$ .

180

$180$

180

$180$

Step 5.2.8

The period of the

tan (θ)

$\tan (θ)$ function is

180

$180$ so values will repeat every

180

$180$ degrees in both directions.

θ = 18.43494882 + 180 n, 198.43494882 + 180 n

$θ = 18.43494882 + 180 n, 198.43494882 + 180 n$ , for any integer

n

$n$

θ = 18.43494882 + 180 n, 198.43494882 + 180 n

$θ = 18.43494882 + 180 n, 198.43494882 + 180 n$ , for any integer

n

$n$

θ = 18.43494882 + 180 n, 198.43494882 + 180 n

$θ = 18.43494882 + 180 n, 198.43494882 + 180 n$ , for any integer

n

$n$

Step 6

Set

2 tan (θ) - 1

$2 \tan (θ) - 1$ equal to

0

$0$ and solve for

θ

$θ$ .

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Step 6.1

Set

2 tan (θ) - 1

$2 \tan (θ) - 1$ equal to

0

$0$ .

2 tan (θ) - 1 = 0

$2 \tan (θ) - 1 = 0$

Step 6.2

Solve

2 tan (θ) - 1 = 0

$2 \tan (θ) - 1 = 0$ for

θ

$θ$ .

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Step 6.2.1

Add

1

$1$ to both sides of the equation.

2 tan (θ) = 1

$2 \tan (θ) = 1$

Step 6.2.2

Divide each term in

2 tan (θ) = 1

$2 \tan (θ) = 1$ by

2

$2$ and simplify.

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Step 6.2.2.1

Divide each term in

2 tan (θ) = 1

$2 \tan (θ) = 1$ by

2

$2$ .

\frac{2 tan (θ)}{2} = \frac{1}{2}

$\frac{2 \tan (θ)}{2} = \frac{1}{2}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of

2

$2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

\frac{2 tan (θ)}{2} = \frac{1}{2}

$\frac{2 \tan (θ)}{2} = \frac{1}{2}$

Step 6.2.2.2.1.2

Divide

tan (θ)

$\tan (θ)$ by

1

$1$ .

tan (θ) = \frac{1}{2}

$\tan (θ) = \frac{1}{2}$

tan (θ) = \frac{1}{2}

$\tan (θ) = \frac{1}{2}$

tan (θ) = \frac{1}{2}

$\tan (θ) = \frac{1}{2}$

tan (θ) = \frac{1}{2}

$\tan (θ) = \frac{1}{2}$

Step 6.2.3

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

θ

$θ$ from inside the tangent.

θ = arctan (\frac{1}{2})

$θ = \arctan (\frac{1}{2})$

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

Evaluate

arctan (\frac{1}{2})

$\arctan (\frac{1}{2})$ .

θ = 26.56505117

$θ = 26.56505117$

θ = 26.56505117

$θ = 26.56505117$

Step 6.2.5

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

180

$180$ to find the solution in the fourth quadrant.

θ = 180 + 26.56505117

$θ = 180 + 26.56505117$

Step 6.2.6

Add

180

$180$ and

26.56505117

$26.56505117$ .

θ = 206.56505117

$θ = 206.56505117$

Step 6.2.7

Find the period of

tan (θ)

$\tan (θ)$ .

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Step 6.2.7.1

The period of the function can be calculated using

\frac{180}{| b |}

$\frac{180}{| b |}$ .

\frac{180}{| b |}

$\frac{180}{| b |}$

Step 6.2.7.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{180}{| 1 |}

$\frac{180}{| 1 |}$

Step 6.2.7.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{180}{1}

$\frac{180}{1}$

Step 6.2.7.4

Divide

180

$180$ by

1

$1$ .

180

$180$

180

$180$

Step 6.2.8

The period of the

tan (θ)

$\tan (θ)$ function is

180

$180$ so values will repeat every

180

$180$ degrees in both directions.

θ = 26.56505117 + 180 n, 206.56505117 + 180 n

$θ = 26.56505117 + 180 n, 206.56505117 + 180 n$ , for any integer

n

$n$

θ = 26.56505117 + 180 n, 206.56505117 + 180 n

$θ = 26.56505117 + 180 n, 206.56505117 + 180 n$ , for any integer

n

$n$

θ = 26.56505117 + 180 n, 206.56505117 + 180 n

$θ = 26.56505117 + 180 n, 206.56505117 + 180 n$ , for any integer

n

$n$

Step 7

The final solution is all the values that make

(3 tan (θ) - 1) (2 tan (θ) - 1) = 0

$(3 \tan (θ) - 1) (2 \tan (θ) - 1) = 0$ true.

θ = 18.43494882 + 180 n, 198.43494882 + 180 n, 26.56505117 + 180 n, 206.56505117 + 180 n

$θ = 18.43494882 + 180 n, 198.43494882 + 180 n, 26.56505117 + 180 n, 206.56505117 + 180 n$ , for any integer

n

$n$

Step 8

Consolidate the answers.

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Step 8.1

Consolidate

18.43494882 + 180 n

$18.43494882 + 180 n$ and

198.43494882 + 180 n

$198.43494882 + 180 n$ to

18.43494882 + 180 n

$18.43494882 + 180 n$ .

θ = 18.43494882 + 180 n, 26.56505117 + 180 n, 206.56505117 + 180 n

$θ = 18.43494882 + 180 n, 26.56505117 + 180 n, 206.56505117 + 180 n$ , for any integer

n

$n$

Step 8.2

Consolidate

26.56505117 + 180 n

$26.56505117 + 180 n$ and

206.56505117 + 180 n

$206.56505117 + 180 n$ to

26.56505117 + 180 n

$26.56505117 + 180 n$ .

θ = 18.43494882 + 180 n, 26.56505117 + 180 n

$θ = 18.43494882 + 180 n, 26.56505117 + 180 n$ , for any integer

n

$n$

θ = 18.43494882 + 180 n, 26.56505117 + 180 n

$θ = 18.43494882 + 180 n, 26.56505117 + 180 n$ , for any integer

n

$n$