Precalculus Examples

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

Step 1

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

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

Step 2

Simplify the right side.

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Step 2.1

The exact value of $arccot (\sqrt{3})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

Step 3

The cotangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = π + \frac{π}{6}$

Step 4

Simplify $π + \frac{π}{6}$ .

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Step 4.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

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

Step 4.2

Combine fractions.

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Step 4.2.1

Combine $π$ and $\frac{6}{6}$ .

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

Step 4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 + π}{6}$

Step 4.3

Simplify the numerator.

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Step 4.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π + π}{6}$

Step 4.3.2

Add $6 π$ and $π$ .

$x = \frac{7 π}{6}$

Step 5

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

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Step 5.1

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

$\frac{π}{| b |}$

Step 5.2

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

$\frac{π}{| 1 |}$

Step 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 5.4

Divide $π$ by $1$ .

$π$

Step 6

The period of the $\cot (x)$ function is $π$ so values will repeat every $π$ radians in both directions.

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

Step 7

Consolidate the answers.

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