Precalculus Examples

$5 \cot^{2} (x) - 15 = 0$

Step 1

Add $15$ to both sides of the equation.

$5 \cot^{2} (x) = 15$

Step 2

Divide each term in $5 \cot^{2} (x) = 15$ by $5$ and simplify.

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Step 2.1

Divide each term in $5 \cot^{2} (x) = 15$ by $5$ .

$\frac{5 \cot^{2} (x)}{5} = \frac{15}{5}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $5$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{5 \cot^{2} (x)}{5} = \frac{15}{5}$

Step 2.2.1.2

Divide $\cot^{2} (x)$ by $1$ .

$\cot^{2} (x) = \frac{15}{5}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Divide $15$ by $5$ .

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

Step 3

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

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

Step 4

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

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

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

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

Step 5

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

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

Simplify the right side.

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Step 6.2.1

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

$x = \frac{π}{6}$

Step 6.3

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 6.4

Simplify $π + \frac{π}{6}$ .

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Step 6.4.1

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 6.4.2

Combine fractions.

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Step 6.4.2.1

Combine $π$ and $\frac{6}{6}$ .

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

Step 6.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 + π}{6}$

Step 6.4.3

Simplify the numerator.

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Step 6.4.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π + π}{6}$

Step 6.4.3.2

Add $6 π$ and $π$ .

$x = \frac{7 π}{6}$

Step 6.5

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

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Step 6.5.1

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

$\frac{π}{| b |}$

Step 6.5.2

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

$\frac{π}{| 1 |}$

Step 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 6.5.4

Divide $π$ by $1$ .

$π$

Step 6.6

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

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

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Step 7.1

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

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

Step 7.2

Simplify the right side.

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Step 7.2.1

The exact value of $arccot (- \sqrt{3})$ is $\frac{5 π}{6}$ .

$x = \frac{5 π}{6}$

Step 7.3

The cotangent 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.

$x = \frac{5 π}{6} - π$

Step 7.4

Simplify the expression to find the second solution.

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Step 7.4.1

Add $2 π$ to $\frac{5 π}{6} - π$ .

$x = \frac{5 π}{6} - π + 2 π$

Step 7.4.2

The resulting angle of $\frac{11 π}{6}$ is positive and coterminal with $\frac{5 π}{6} - π$ .

$x = \frac{11 π}{6}$

Step 7.5

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

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Step 7.5.1

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

$\frac{π}{| b |}$

Step 7.5.2

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

$\frac{π}{| 1 |}$

Step 7.5.3

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

$\frac{π}{1}$

Step 7.5.4

Divide $π$ by $1$ .

$π$

Step 7.6

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

$x = \frac{5 π}{6} + π n, \frac{11 π}{6} + π n$ , for any integer $n$

Step 8

List all of the solutions.

$x = \frac{π}{6} + π n, \frac{7 π}{6} + π n, \frac{5 π}{6} + π n, \frac{11 π}{6} + π n$ , for any integer $n$

Step 9

Consolidate the solutions.

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Step 9.1

Consolidate $\frac{π}{6} + π n$ and $\frac{7 π}{6} + π n$ to $\frac{π}{6} + π n$ .

$x = \frac{π}{6} + π n, \frac{5 π}{6} + π n, \frac{11 π}{6} + π n$ , for any integer $n$

Step 9.2

Consolidate $\frac{5 π}{6} + π n$ and $\frac{11 π}{6} + π n$ to $\frac{5 π}{6} + π n$ .

$x = \frac{π}{6} + π n, \frac{5 π}{6} + π n$ , for any integer $n$

Step 10