Trigonometry Examples

\cot (x) = \sqrt{3}

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

Step 1

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

x

$x$ from inside the cotangent.

x = arccot (\sqrt{3})

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

Step 2

Simplify the right side.

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Step 2.1

The exact value of

arccot (\sqrt{3})

$arccot (\sqrt{3})$ is

\frac{π}{6}

$\frac{π}{6}$ .

x = \frac{π}{6}

$x = \frac{π}{6}$

x = \frac{π}{6}

$x = \frac{π}{6}$

Step 3

The cotangent 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}{6}

$x = π + \frac{π}{6}$

Step 4

Simplify

π + \frac{π}{6}

$π + \frac{π}{6}$ .

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Step 4.1

To write

π

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

\frac{6}{6}

$\frac{6}{6}$ .

x = π \cdot \frac{6}{6} + \frac{π}{6}

$x = π \cdot \frac{6}{6} + \frac{π}{6}$

Step 4.2

Combine fractions.

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Step 4.2.1

Combine

π

$π$ and

\frac{6}{6}

$\frac{6}{6}$ .

x = \frac{π \cdot 6}{6} + \frac{π}{6}

$x = \frac{π \cdot 6}{6} + \frac{π}{6}$

Step 4.2.2

Combine the numerators over the common denominator.

x = \frac{π \cdot 6 + π}{6}

$x = \frac{π \cdot 6 + π}{6}$

x = \frac{π \cdot 6 + π}{6}

$x = \frac{π \cdot 6 + π}{6}$

Step 4.3

Simplify the numerator.

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Step 4.3.1

Move

6

$6$ to the left of

π

$π$ .

x = \frac{6 \cdot π + π}{6}

$x = \frac{6 \cdot π + π}{6}$

Step 4.3.2

Add

6 π

$6 π$ and

π

$π$ .

x = \frac{7 π}{6}

$x = \frac{7 π}{6}$

x = \frac{7 π}{6}

$x = \frac{7 π}{6}$

x = \frac{7 π}{6}

$x = \frac{7 π}{6}$

Step 5

Find the period of

\cot (x)

$\cot (x)$ .

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Step 5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 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 5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 6

The period of the

\cot (x)

$\cot (x)$ function is

π

$π$ so values will repeat every

π

$π$ radians in both directions.

x = \frac{π}{6} + π n, \frac{7 π}{6} + π n

$x = \frac{π}{6} + π n, \frac{7 π}{6} + π n$ , for any integer

n

$n$

Step 7

Consolidate the answers.

x = \frac{π}{6} + π n

$x = \frac{π}{6} + π n$ , for any integer

n

$n$