Algebra Examples

\tan^{2} (x) = 3

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

Step 1

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

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

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

Step 2

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

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Step 2.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

\tan (x) = \sqrt{3}

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

Step 2.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

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

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

Step 2.3

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

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

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

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

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

Step 3

Set up each of the solutions to solve for

x

$x$ .

\tan (x) = \sqrt{3}

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

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

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

Step 4

Solve for

x

$x$ in

\tan (x) = \sqrt{3}

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

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Step 4.1

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

x

$x$ from inside the tangent.

x = \arctan (\sqrt{3})

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

Step 4.2

Simplify the right side.

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Step 4.2.1

The exact value of

\arctan (\sqrt{3})

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

\frac{π}{3}

$\frac{π}{3}$ .

x = \frac{π}{3}

$x = \frac{π}{3}$

x = \frac{π}{3}

$x = \frac{π}{3}$

Step 4.3

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

π

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

x = π + \frac{π}{3}

$x = π + \frac{π}{3}$

Step 4.4

Simplify

π + \frac{π}{3}

$π + \frac{π}{3}$ .

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Step 4.4.1

To write

π

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

\frac{3}{3}

$\frac{3}{3}$ .

x = π \cdot \frac{3}{3} + \frac{π}{3}

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

Step 4.4.2

Combine fractions.

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Step 4.4.2.1

Combine

π

$π$ and

\frac{3}{3}

$\frac{3}{3}$ .

x = \frac{π \cdot 3}{3} + \frac{π}{3}

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

Step 4.4.2.2

Combine the numerators over the common denominator.

x = \frac{π \cdot 3 + π}{3}

$x = \frac{π \cdot 3 + π}{3}$

x = \frac{π \cdot 3 + π}{3}

$x = \frac{π \cdot 3 + π}{3}$

Step 4.4.3

Simplify the numerator.

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Step 4.4.3.1

Move

3

$3$ to the left of

π

$π$ .

x = \frac{3 \cdot π + π}{3}

$x = \frac{3 \cdot π + π}{3}$

Step 4.4.3.2

Add

3 π

$3 π$ and

π

$π$ .

x = \frac{4 π}{3}

$x = \frac{4 π}{3}$

x = \frac{4 π}{3}

$x = \frac{4 π}{3}$

x = \frac{4 π}{3}

$x = \frac{4 π}{3}$

Step 4.5

Find the period of

\tan (x)

$\tan (x)$ .

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Step 4.5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 4.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 4.5.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{π}{1}

$\frac{π}{1}$

Step 4.5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 4.6

The period of the

\tan (x)

$\tan (x)$ function is

π

$π$ so values will repeat every

π

$π$ radians in both directions.

x = \frac{π}{3} + π n, \frac{4 π}{3} + π n

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

n

$n$

x = \frac{π}{3} + π n, \frac{4 π}{3} + π n

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

n

$n$

Step 5

Solve for

x

$x$ in

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

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

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Step 5.1

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

x

$x$ from inside the tangent.

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

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

Step 5.2

Simplify the right side.

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Step 5.2.1

The exact value of

\arctan (- \sqrt{3})

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

- \frac{π}{3}

$- \frac{π}{3}$ .

x = - \frac{π}{3}

$x = - \frac{π}{3}$

x = - \frac{π}{3}

$x = - \frac{π}{3}$

Step 5.3

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

π

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

x = - \frac{π}{3} - π

$x = - \frac{π}{3} - π$

Step 5.4

Simplify the expression to find the second solution.

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Step 5.4.1

Add

2 π

$2 π$ to

- \frac{π}{3} - π

$- \frac{π}{3} - π$ .

x = - \frac{π}{3} - π + 2 π

$x = - \frac{π}{3} - π + 2 π$

Step 5.4.2

The resulting angle of

\frac{2 π}{3}

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

- \frac{π}{3} - π

$- \frac{π}{3} - π$ .

x = \frac{2 π}{3}

$x = \frac{2 π}{3}$

x = \frac{2 π}{3}

$x = \frac{2 π}{3}$

Step 5.5

Find the period of

\tan (x)

$\tan (x)$ .

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Step 5.5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 5.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 5.5.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{π}{1}

$\frac{π}{1}$

Step 5.5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 5.6

Add

π

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

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Step 5.6.1

Add

π

$π$ to

- \frac{π}{3}

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

- \frac{π}{3} + π

$- \frac{π}{3} + π$

Step 5.6.2

To write

π

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

\frac{3}{3}

$\frac{3}{3}$ .

π \cdot \frac{3}{3} - \frac{π}{3}

$π \cdot \frac{3}{3} - \frac{π}{3}$

Step 5.6.3

Combine fractions.

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Step 5.6.3.1

Combine

π

$π$ and

\frac{3}{3}

$\frac{3}{3}$ .

\frac{π \cdot 3}{3} - \frac{π}{3}

$\frac{π \cdot 3}{3} - \frac{π}{3}$

Step 5.6.3.2

Combine the numerators over the common denominator.

\frac{π \cdot 3 - π}{3}

$\frac{π \cdot 3 - π}{3}$

\frac{π \cdot 3 - π}{3}

$\frac{π \cdot 3 - π}{3}$

Step 5.6.4

Simplify the numerator.

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Step 5.6.4.1

Move

3

$3$ to the left of

π

$π$ .

\frac{3 \cdot π - π}{3}

$\frac{3 \cdot π - π}{3}$

Step 5.6.4.2

Subtract

π

$π$ from

3 π

$3 π$ .

\frac{2 π}{3}

$\frac{2 π}{3}$

\frac{2 π}{3}

$\frac{2 π}{3}$

Step 5.6.5

List the new angles.

x = \frac{2 π}{3}

$x = \frac{2 π}{3}$

x = \frac{2 π}{3}

$x = \frac{2 π}{3}$

Step 5.7

The period of the

\tan (x)

$\tan (x)$ function is

π

$π$ so values will repeat every

π

$π$ radians in both directions.

x = \frac{2 π}{3} + π n, \frac{2 π}{3} + π n

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

n

$n$

x = \frac{2 π}{3} + π n, \frac{2 π}{3} + π n

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

n

$n$

Step 6

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

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

n

$n$

Step 7

Consolidate the solutions.

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Step 7.1

Consolidate

\frac{π}{3} + π n

$\frac{π}{3} + π n$ and

\frac{4 π}{3} + π n

$\frac{4 π}{3} + π n$ to

\frac{π}{3} + π n

$\frac{π}{3} + π n$ .

x = \frac{π}{3} + π n, \frac{2 π}{3} + π n, \frac{2 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

$n$

Step 7.2

Consolidate

\frac{2 π}{3} + π n

$\frac{2 π}{3} + π n$ and

\frac{2 π}{3} + π n

$\frac{2 π}{3} + π n$ to

\frac{2 π}{3} + π n

$\frac{2 π}{3} + π n$ .

x = \frac{π}{3} + π n, \frac{2 π}{3} + π n

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

n

$n$

x = \frac{π}{3} + π n, \frac{2 π}{3} + π n

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

n

$n$