Trigonometry Examples

$8 \tan^{2} (θ) + 10 \tan (θ) + 10 = 7$

Step 1

Subtract

$7$ from both sides of the equation.

$8 \tan^{2} (θ) + 10 \tan (θ) + 10 - 7 = 0$

Step 2

Subtract

$7$ from

$10$ .

$8 \tan^{2} (θ) + 10 \tan (θ) + 3 = 0$

Step 3

Factor by grouping.

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Step 3.1

For a polynomial of the form

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

$a \cdot c = 8 \cdot 3 = 24$ and whose sum is

$b = 10$ .

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Step 3.1.1

Factor

$10$ out of

$10 \tan (θ)$ .

$8 \tan^{2} (θ) + 10 \tan (θ) + 3 = 0$

Step 3.1.2

Rewrite

$10$ as

$4$ plus

$6$

$8 \tan^{2} (θ) + (4 + 6) \tan (θ) + 3 = 0$

Step 3.1.3

Apply the distributive property.

$8 \tan^{2} (θ) + 4 \tan (θ) + 6 \tan (θ) + 3 = 0$

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

$8 \tan^{2} (θ) + 4 \tan (θ) + 6 \tan (θ) + 3 = 0$

Step 3.2.2

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

$4 \tan (θ) (2 \tan (θ) + 1) + 3 (2 \tan (θ) + 1) = 0$

Step 3.3

Factor the polynomial by factoring out the greatest common factor,

$2 \tan (θ) + 1$ .

$(2 \tan (θ) + 1) (4 \tan (θ) + 3) = 0$

Step 4

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

$0$ , the entire expression will be equal to

$0$ .

$2 \tan (θ) + 1 = 0$

$4 \tan (θ) + 3 = 0$

Step 5

Set

$2 \tan (θ) + 1$ equal to

$0$ and solve for

$θ$ .

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Step 5.1

Set

$2 \tan (θ) + 1$ equal to

$0$ .

$2 \tan (θ) + 1 = 0$

Step 5.2

Solve

$2 \tan (θ) + 1 = 0$ for

$θ$ .

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Step 5.2.1

Subtract

$1$ from both sides of the equation.

$2 \tan (θ) = - 1$

Step 5.2.2

Divide each term in

$2 \tan (θ) = - 1$ by

$2$ and simplify.

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Step 5.2.2.1

Divide each term in

$2 \tan (θ) = - 1$ by

$2$ .

$\frac{2 \tan (θ)}{2} = \frac{- 1}{2}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of

$2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{2 \tan (θ)}{2} = \frac{- 1}{2}$

Step 5.2.2.2.1.2

Divide

$\tan (θ)$ by

$1$ .

$\tan (θ) = \frac{- 1}{2}$

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

Move the negative in front of the fraction.

$\tan (θ) = - \frac{1}{2}$

Step 5.2.3

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

$θ$ from inside the tangent.

$θ = \arctan (- \frac{1}{2})$

Step 5.2.4

Simplify the right side.

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Step 5.2.4.1

Evaluate

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

$θ = - 26.56505117$

Step 5.2.5

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from

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

$θ = - 26.56505117 - 180$

Step 5.2.6

Simplify the expression to find the second solution.

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Step 5.2.6.1

Add

$360 °$ to

$- 26.56505117 - 180 °$ .

$θ = - 26.56505117 - 180 ° + 360 °$

Step 5.2.6.2

The resulting angle of

$153.43494882 °$ is positive and coterminal with

$- 26.56505117 - 180$ .

$θ = 153.43494882 °$

Step 5.2.7

Find the period of

$\tan (θ)$ .

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Step 5.2.7.1

The period of the function can be calculated using

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

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

Step 5.2.7.2

Replace

$b$ with

$1$ in the formula for period.

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

Step 5.2.7.3

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

$0$ and

$1$ is

$1$ .

$\frac{180}{1}$

Step 5.2.7.4

Divide

$180$ by

$1$ .

$180$

Step 5.2.8

Add

$180$ to every negative angle to get positive angles.

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Step 5.2.8.1

Add

$180$ to

$- 26.56505117$ to find the positive angle.

$- 26.56505117 + 180$

Step 5.2.8.2

Subtract

$26.56505117$ from

$180$ .

$153.43494882$

Step 5.2.8.3

List the new angles.

$θ = 153.43494882$

Step 5.2.9

The period of the

$\tan (θ)$ function is

$180$ so values will repeat every

$180$ degrees in both directions.

$θ = 153.43494882 + 180 n, 153.43494882 + 180 n$ , for any integer

$n$

$θ = 153.43494882 + 180 n, 153.43494882 + 180 n$ , for any integer

$n$

$θ = 153.43494882 + 180 n, 153.43494882 + 180 n$ , for any integer

$n$

Step 6

Set

$4 \tan (θ) + 3$ equal to

$0$ and solve for

$θ$ .

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Step 6.1

Set

$4 \tan (θ) + 3$ equal to

$0$ .

$4 \tan (θ) + 3 = 0$

Step 6.2

Solve

$4 \tan (θ) + 3 = 0$ for

$θ$ .

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Step 6.2.1

Subtract

$3$ from both sides of the equation.

$4 \tan (θ) = - 3$

Step 6.2.2

Divide each term in

$4 \tan (θ) = - 3$ by

$4$ and simplify.

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Step 6.2.2.1

Divide each term in

$4 \tan (θ) = - 3$ by

$4$ .

$\frac{4 \tan (θ)}{4} = \frac{- 3}{4}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of

$4$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{4 \tan (θ)}{4} = \frac{- 3}{4}$

Step 6.2.2.2.1.2

Divide

$\tan (θ)$ by

$1$ .

$\tan (θ) = \frac{- 3}{4}$

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Move the negative in front of the fraction.

$\tan (θ) = - \frac{3}{4}$

Step 6.2.3

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

$θ$ from inside the tangent.

$θ = \arctan (- \frac{3}{4})$

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

Evaluate

$\arctan (- \frac{3}{4})$ .

$θ = - 36.86989764$

Step 6.2.5

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from

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

$θ = - 36.86989764 - 180$

Step 6.2.6

Simplify the expression to find the second solution.

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Step 6.2.6.1

Add

$360 °$ to

$- 36.86989764 - 180 °$ .

$θ = - 36.86989764 - 180 ° + 360 °$

Step 6.2.6.2

The resulting angle of

$143.13010235 °$ is positive and coterminal with

$- 36.86989764 - 180$ .

$θ = 143.13010235 °$

Step 6.2.7

Find the period of

$\tan (θ)$ .

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Step 6.2.7.1

The period of the function can be calculated using

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

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

Step 6.2.7.2

Replace

$b$ with

$1$ in the formula for period.

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

Step 6.2.7.3

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

$0$ and

$1$ is

$1$ .

$\frac{180}{1}$

Step 6.2.7.4

Divide

$180$ by

$1$ .

$180$

Step 6.2.8

Add

$180$ to every negative angle to get positive angles.

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Step 6.2.8.1

Add

$180$ to

$- 36.86989764$ to find the positive angle.

$- 36.86989764 + 180$

Step 6.2.8.2

Subtract

$36.86989764$ from

$180$ .

$143.13010235$

Step 6.2.8.3

List the new angles.

$θ = 143.13010235$

Step 6.2.9

The period of the

$\tan (θ)$ function is

$180$ so values will repeat every

$180$ degrees in both directions.

$θ = 143.13010235 + 180 n, 143.13010235 + 180 n$ , for any integer

$n$

$θ = 143.13010235 + 180 n, 143.13010235 + 180 n$ , for any integer

$n$

$θ = 143.13010235 + 180 n, 143.13010235 + 180 n$ , for any integer

$n$

Step 7

The final solution is all the values that make

$(2 \tan (θ) + 1) (4 \tan (θ) + 3) = 0$ true.

$θ = 153.43494882 + 180 n, 143.13010235 + 180 n$ , for any integer

$n$