Trigonometry Examples

{tan}^{5} (x) - 9 tan (x) = 0

$\tan^{5} (x) - 9 \tan (x) = 0$

Step 1

Factor

$\tan^{5} (x) - 9 \tan (x)$ .

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Step 1.1

Factor

$\tan (x)$ out of

$\tan^{5} (x) - 9 \tan (x)$ .

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Step 1.1.1

Factor

$\tan (x)$ out of

$\tan^{5} (x)$ .

$\tan (x) \tan^{4} (x) - 9 \tan (x) = 0$

Step 1.1.2

Factor

$\tan (x)$ out of

$- 9 \tan (x)$ .

$\tan (x) \tan^{4} (x) + \tan (x) \cdot - 9 = 0$

Step 1.1.3

Factor

$\tan (x)$ out of

$\tan (x) \tan^{4} (x) + \tan (x) \cdot - 9$ .

$\tan (x) (\tan^{4} (x) - 9) = 0$

Step 1.2

Rewrite

$\tan^{4} (x)$ as

${(\tan^{2} (x))}^{2}$ .

$\tan (x) ({(\tan^{2} (x))}^{2} - 9) = 0$

Step 1.3

Rewrite

$9$ as

$3^{2}$ .

$\tan (x) ({(\tan^{2} (x))}^{2} - 3^{2}) = 0$

Step 1.4

Factor.

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Step 1.4.1

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = \tan^{2} (x)$ and

$b = 3$ .

$\tan (x) ((\tan^{2} (x) + 3) (\tan^{2} (x) - 3)) = 0$

Step 1.4.2

Remove unnecessary parentheses.

$\tan (x) (\tan^{2} (x) + 3) (\tan^{2} (x) - 3) = 0$

Step 2

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

$0$ , the entire expression will be equal to

$0$ .

$\tan (x) = 0$

$\tan^{2} (x) + 3 = 0$

$\tan^{2} (x) - 3 = 0$

Step 3

Set

$\tan (x)$ equal to

$0$ and solve for

$x$ .

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Step 3.1

Set

$\tan (x)$ equal to

$0$ .

$\tan (x) = 0$

Step 3.2

Solve

$\tan (x) = 0$ for

$x$ .

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Step 3.2.1

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

$x$ from inside the tangent.

$x = \arctan (0)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

The exact value of

$\arctan (0)$ is

$0$ .

$x = 0$

Step 3.2.3

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

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

$x = π + 0$

Step 3.2.4

Add

$π$ and

$0$ .

$x = π$

Step 3.2.5

Find the period of

$\tan (x)$ .

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Step 3.2.5.1

The period of the function can be calculated using

$\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 3.2.5.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 3.2.5.3

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

$0$ and

$1$ is

$1$ .

$\frac{π}{1}$

Step 3.2.5.4

Divide

$π$ by

$1$ .

$π$

Step 3.2.6

The period of the

$\tan (x)$ function is

$π$ so values will repeat every

$π$ radians in both directions.

$x = π n, π + π n$ , for any integer

$n$

$x = π n, π + π n$ , for any integer

$n$

$x = π n, π + π n$ , for any integer

$n$

Step 4

Set

$\tan^{2} (x) + 3$ equal to

$0$ and solve for

$x$ .

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Step 4.1

Set

$\tan^{2} (x) + 3$ equal to

$0$ .

$\tan^{2} (x) + 3 = 0$

Step 4.2

Solve

$\tan^{2} (x) + 3 = 0$ for

$x$ .

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Step 4.2.1

Subtract

$3$ from both sides of the equation.

$\tan^{2} (x) = - 3$

Step 4.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\tan (x) = \pm \sqrt{- 3}$

Step 4.2.3

Simplify

$\pm \sqrt{- 3}$ .

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Step 4.2.3.1

Rewrite

$- 3$ as

$- 1 (3)$ .

$\tan (x) = \pm \sqrt{- 1 (3)}$

Step 4.2.3.2

Rewrite

$\sqrt{- 1 (3)}$ as

$\sqrt{- 1} \cdot \sqrt{3}$ .

$\tan (x) = \pm \sqrt{- 1} \cdot \sqrt{3}$

Step 4.2.3.3

Rewrite

$\sqrt{- 1}$ as

$i$ .

$\tan (x) = \pm i \sqrt{3}$

Step 4.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.2.4.1

First, use the positive value of the

$\pm$ to find the first solution.

$\tan (x) = i \sqrt{3}$

Step 4.2.4.2

Next, use the negative value of the

$\pm$ to find the second solution.

$\tan (x) = - i \sqrt{3}$

Step 4.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$\tan (x) = i \sqrt{3}, - i \sqrt{3}$

Step 4.2.5

Set up each of the solutions to solve for

$x$ .

$\tan (x) = i \sqrt{3}$

$\tan (x) = - i \sqrt{3}$

Step 4.2.6

Solve for

$x$ in

$\tan (x) = i \sqrt{3}$ .

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Step 4.2.6.1

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

$x$ from inside the tangent.

$x = \arctan (i \sqrt{3})$

Step 4.2.6.2

The inverse tangent of

$\arctan (i \sqrt{3})$ is undefined.

Undefined

Step 4.2.7

Solve for

$x$ in

$\tan (x) = - i \sqrt{3}$ .

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Step 4.2.7.1

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

$x$ from inside the tangent.

$x = \arctan (- i \sqrt{3})$

Step 4.2.7.2

The inverse tangent of

$\arctan (- i \sqrt{3})$ is undefined.

Undefined

Step 4.2.8

List all of the solutions.

No solution

Step 5

Set

$\tan^{2} (x) - 3$ equal to

$0$ and solve for

$x$ .

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Step 5.1

Set

$\tan^{2} (x) - 3$ equal to

$0$ .

$\tan^{2} (x) - 3 = 0$

Step 5.2

Solve

$\tan^{2} (x) - 3 = 0$ for

$x$ .

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Step 5.2.1

Add

$3$ to both sides of the equation.

$\tan^{2} (x) = 3$

Step 5.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\tan (x) = \pm \sqrt{3}$

Step 5.2.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.2.3.1

First, use the positive value of the

$\pm$ to find the first solution.

$\tan (x) = \sqrt{3}$

Step 5.2.3.2

Next, use the negative value of the

$\pm$ to find the second solution.

$\tan (x) = - \sqrt{3}$

Step 5.2.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$\tan (x) = \sqrt{3}, - \sqrt{3}$

Step 5.2.4

Set up each of the solutions to solve for

$x$ .

$\tan (x) = \sqrt{3}$

$\tan (x) = - \sqrt{3}$

Step 5.2.5

Solve for

$x$ in

$\tan (x) = \sqrt{3}$ .

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Step 5.2.5.1

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

$x$ from inside the tangent.

$x = \arctan (\sqrt{3})$

Step 5.2.5.2

Simplify the right side.

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Step 5.2.5.2.1

The exact value of

$\arctan (\sqrt{3})$ is

$\frac{π}{3}$ .

$x = \frac{π}{3}$

Step 5.2.5.3

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

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

$x = π + \frac{π}{3}$

Step 5.2.5.4

Simplify

$π + \frac{π}{3}$ .

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Step 5.2.5.4.1

To write

$π$ as a fraction with a common denominator, multiply by

$\frac{3}{3}$ .

$x = π \cdot \frac{3}{3} + \frac{π}{3}$

Step 5.2.5.4.2

Combine fractions.

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Step 5.2.5.4.2.1

Combine

$π$ and

$\frac{3}{3}$ .

$x = \frac{π \cdot 3}{3} + \frac{π}{3}$

Step 5.2.5.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 3 + π}{3}$

Step 5.2.5.4.3

Simplify the numerator.

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Step 5.2.5.4.3.1

Move

$3$ to the left of

$π$ .

$x = \frac{3 \cdot π + π}{3}$

Step 5.2.5.4.3.2

Add

$3 π$ and

$π$ .

$x = \frac{4 π}{3}$

Step 5.2.5.5

Find the period of

$\tan (x)$ .

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Step 5.2.5.5.1

The period of the function can be calculated using

$\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 5.2.5.5.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 5.2.5.5.3

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

$0$ and

$1$ is

$1$ .

$\frac{π}{1}$

Step 5.2.5.5.4

Divide

$π$ by

$1$ .

$π$

Step 5.2.5.6

The period of the

$\tan (x)$ function is

$π$ so values will repeat every

$π$ radians in both directions.

$x = \frac{π}{3} + π n, \frac{4 π}{3} + π n$ , for any integer

$n$

$x = \frac{π}{3} + π n, \frac{4 π}{3} + π n$ , for any integer

$n$

Step 5.2.6

Solve for

$x$ in

$\tan (x) = - \sqrt{3}$ .

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Step 5.2.6.1

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

$x$ from inside the tangent.

$x = \arctan (- \sqrt{3})$

Step 5.2.6.2

Simplify the right side.

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Step 5.2.6.2.1

The exact value of

$\arctan (- \sqrt{3})$ is

$- \frac{π}{3}$ .

$x = - \frac{π}{3}$

Step 5.2.6.3

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

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

$x = - \frac{π}{3} - π$

Step 5.2.6.4

Simplify the expression to find the second solution.

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Step 5.2.6.4.1

Add

$2 π$ to

$- \frac{π}{3} - π$ .

$x = - \frac{π}{3} - π + 2 π$

Step 5.2.6.4.2

The resulting angle of

$\frac{2 π}{3}$ is positive and coterminal with

$- \frac{π}{3} - π$ .

$x = \frac{2 π}{3}$

Step 5.2.6.5

Find the period of

$\tan (x)$ .

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Step 5.2.6.5.1

The period of the function can be calculated using

$\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 5.2.6.5.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 5.2.6.5.3

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

$0$ and

$1$ is

$1$ .

$\frac{π}{1}$

Step 5.2.6.5.4

Divide

$π$ by

$1$ .

$π$

Step 5.2.6.6

Add

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

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Step 5.2.6.6.1

Add

$π$ to

$- \frac{π}{3}$ to find the positive angle.

$- \frac{π}{3} + π$

Step 5.2.6.6.2

To write

$π$ as a fraction with a common denominator, multiply by

$\frac{3}{3}$ .

$π \cdot \frac{3}{3} - \frac{π}{3}$

Step 5.2.6.6.3

Combine fractions.

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Step 5.2.6.6.3.1

Combine

$π$ and

$\frac{3}{3}$ .

$\frac{π \cdot 3}{3} - \frac{π}{3}$

Step 5.2.6.6.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 3 - π}{3}$

Step 5.2.6.6.4

Simplify the numerator.

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Step 5.2.6.6.4.1

Move

$3$ to the left of

$π$ .

$\frac{3 \cdot π - π}{3}$

Step 5.2.6.6.4.2

Subtract

$π$ from

$3 π$ .

$\frac{2 π}{3}$

Step 5.2.6.6.5

List the new angles.

$x = \frac{2 π}{3}$

Step 5.2.6.7

The period of the

$\tan (x)$ function is

$π$ so values will repeat every

$π$ radians in both directions.

$x = \frac{2 π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer

$n$

$x = \frac{2 π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer

$n$

Step 5.2.7

List all of the solutions.

$x = \frac{π}{3} + π n, \frac{4 π}{3} + π n, \frac{2 π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer

$n$

Step 5.2.8

Consolidate the solutions.

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Step 5.2.8.1

Consolidate

$\frac{π}{3} + π n$ and

$\frac{4 π}{3} + π n$ to

$\frac{π}{3} + π n$ .

$x = \frac{π}{3} + π n, \frac{2 π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer

$n$

Step 5.2.8.2

Consolidate

$\frac{2 π}{3} + π n$ and

$\frac{2 π}{3} + π n$ to

$\frac{2 π}{3} + π n$ .

$x = \frac{π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer

$n$

$x = \frac{π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer

$n$

$x = \frac{π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer

$n$

$x = \frac{π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer

$n$

Step 6

The final solution is all the values that make

$\tan (x) (\tan^{2} (x) + 3) (\tan^{2} (x) - 3) = 0$ true.

$x = π n, π + π n, \frac{π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer

$n$

Step 7

Consolidate the answers.

$x = \frac{π n}{3}$ , for any integer

$n$