Trigonometry Examples

$\frac{1}{3} \cdot \tan^{3} (x) - \tan (x) = 0$

Step 1

Substitute $u$ for $\tan (x)$ .

$\frac{1}{3} {(u)}^{3} - (u) = 0$

Step 2

Combine $\frac{1}{3}$ and $u^{3}$ .

$\frac{u^{3}}{3} - u = 0$

Step 3

Multiply each term in $\frac{u^{3}}{3} - u = 0$ by $3$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{u^{3}}{3} - u = 0$ by $3$ .

$\frac{u^{3}}{3} \cdot 3 - u \cdot 3 = 0 \cdot 3$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Cancel the common factor of $3$ .

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Step 3.2.1.1.1

Cancel the common factor.

$\frac{u^{3}}{3} \cdot 3 - u \cdot 3 = 0 \cdot 3$

Step 3.2.1.1.2

Rewrite the expression.

$u^{3} - u \cdot 3 = 0 \cdot 3$

Step 3.2.1.2

Multiply $3$ by $- 1$ .

$u^{3} - 3 u = 0 \cdot 3$

Step 3.3

Simplify the right side.

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Step 3.3.1

Multiply $0$ by $3$ .

$u^{3} - 3 u = 0$

Step 4

Factor $u$ out of $u^{3} - 3 u$ .

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Step 4.1

Factor $u$ out of $u^{3}$ .

$u \cdot u^{2} - 3 u = 0$

Step 4.2

Factor $u$ out of $- 3 u$ .

$u \cdot u^{2} + u \cdot - 3 = 0$

Step 4.3

Factor $u$ out of $u \cdot u^{2} + u \cdot - 3$ .

$u (u^{2} - 3) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$u = 0$

$u^{2} - 3 = 0$

Step 6

Set $u$ equal to $0$ .

$u = 0$

Step 7

Set $u^{2} - 3$ equal to $0$ and solve for $u$ .

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Step 7.1

Set $u^{2} - 3$ equal to $0$ .

$u^{2} - 3 = 0$

Step 7.2

Solve $u^{2} - 3 = 0$ for $u$ .

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Step 7.2.1

Add $3$ to both sides of the equation.

$u^{2} = 3$

Step 7.2.2

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

$u = \pm \sqrt{3}$

Step 7.2.3

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

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Step 7.2.3.1

First, use the positive value of the $\pm$ to find the first solution.

$u = \sqrt{3}$

Step 7.2.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$u = - \sqrt{3}$

Step 7.2.3.3

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

$u = \sqrt{3}, - \sqrt{3}$

Step 8

The final solution is all the values that make $u (u^{2} - 3) = 0$ true.

$u = 0, \sqrt{3}, - \sqrt{3}$

Step 9

Substitute $\tan (x)$ for $u$ .

$\tan (x) = 0, \sqrt{3}, - \sqrt{3}$

Step 10

Set up each of the solutions to solve for $x$ .

$\tan (x) = 0$

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

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

Step 11

Solve for $x$ in $\tan (x) = 0$ .

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Step 11.1

Take the inverse tangent of both sides of the equation to extract $x$ from inside the tangent.

$x = \arctan (0)$

Step 11.2

Simplify the right side.

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Step 11.2.1

The exact value of $\arctan (0)$ is $0$ .

$x = 0$

Step 11.3

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 11.4

Add $π$ and $0$ .

$x = π$

Step 11.5

Find the period of $\tan (x)$ .

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Step 11.5.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 11.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 11.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 11.5.4

Divide $π$ by $1$ .

$π$

Step 11.6

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 $n$

Step 12

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

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Step 12.1

Take the inverse tangent of both sides of the equation to extract $x$ from inside the tangent.

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

Step 12.2

Simplify the right side.

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Step 12.2.1

The exact value of $\arctan (\sqrt{3})$ is $\frac{π}{3}$ .

$x = \frac{π}{3}$

Step 12.3

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 12.4

Simplify $π + \frac{π}{3}$ .

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Step 12.4.1

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 12.4.2

Combine fractions.

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Step 12.4.2.1

Combine $π$ and $\frac{3}{3}$ .

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

Step 12.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 3 + π}{3}$

Step 12.4.3

Simplify the numerator.

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Step 12.4.3.1

Move $3$ to the left of $π$ .

$x = \frac{3 \cdot π + π}{3}$

Step 12.4.3.2

Add $3 π$ and $π$ .

$x = \frac{4 π}{3}$

Step 12.5

Find the period of $\tan (x)$ .

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Step 12.5.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 12.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 12.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 12.5.4

Divide $π$ by $1$ .

$π$

Step 12.6

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 $n$

Step 13

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

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Step 13.1

Take the inverse tangent of both sides of the equation to extract $x$ from inside the tangent.

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

Step 13.2

Simplify the right side.

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Step 13.2.1

The exact value of $\arctan (- \sqrt{3})$ is $- \frac{π}{3}$ .

$x = - \frac{π}{3}$

Step 13.3

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 13.4

Simplify the expression to find the second solution.

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Step 13.4.1

Add $2 π$ to $- \frac{π}{3} - π$ .

$x = - \frac{π}{3} - π + 2 π$

Step 13.4.2

The resulting angle of $\frac{2 π}{3}$ is positive and coterminal with $- \frac{π}{3} - π$ .

$x = \frac{2 π}{3}$

Step 13.5

Find the period of $\tan (x)$ .

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Step 13.5.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 13.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 13.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 13.5.4

Divide $π$ by $1$ .

$π$

Step 13.6

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

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Step 13.6.1

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

$- \frac{π}{3} + π$

Step 13.6.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$π \cdot \frac{3}{3} - \frac{π}{3}$

Step 13.6.3

Combine fractions.

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Step 13.6.3.1

Combine $π$ and $\frac{3}{3}$ .

$\frac{π \cdot 3}{3} - \frac{π}{3}$

Step 13.6.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 3 - π}{3}$

Step 13.6.4

Simplify the numerator.

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Step 13.6.4.1

Move $3$ to the left of $π$ .

$\frac{3 \cdot π - π}{3}$

Step 13.6.4.2

Subtract $π$ from $3 π$ .

$\frac{2 π}{3}$

Step 13.6.5

List the new angles.

$x = \frac{2 π}{3}$

Step 13.7

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 $n$

Step 14

List all of the solutions.

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

Step 15

Consolidate the answers.

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