Trigonometry Examples

{cot}^{4} (x) - 4 {cot}^{2} (x) + 3 = 0

$\cot^{4} (x) - 4 \cot^{2} (x) + 3 = 0$

Step 1

Factor the left side of the equation.

Tap for more steps...

Step 1.1

Rewrite

{cot}^{4} (x)

$\cot^{4} (x)$ as

{({cot}^{2} (x))}^{2}

${(\cot^{2} (x))}^{2}$ .

{({cot}^{2} (x))}^{2} - 4 {cot}^{2} (x) + 3 = 0

${(\cot^{2} (x))}^{2} - 4 \cot^{2} (x) + 3 = 0$

Step 1.2

Let

u = {cot}^{2} (x)

$u = \cot^{2} (x)$ . Substitute

u

$u$ for all occurrences of

{cot}^{2} (x)

$\cot^{2} (x)$ .

u^{2} - 4 u + 3 = 0

$u^{2} - 4 u + 3 = 0$

Step 1.3

Factor

u^{2} - 4 u + 3

$u^{2} - 4 u + 3$ using the AC method.

Tap for more steps...

Step 1.3.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

3

$3$ and whose sum is

- 4

$- 4$ .

- 3, - 1

$- 3, - 1$

Step 1.3.2

Write the factored form using these integers.

(u - 3) (u - 1) = 0

$(u - 3) (u - 1) = 0$

(u - 3) (u - 1) = 0

$(u - 3) (u - 1) = 0$

Step 1.4

Replace all occurrences of

u

$u$ with

{cot}^{2} (x)

$\cot^{2} (x)$ .

({cot}^{2} (x) - 3) ({cot}^{2} (x) - 1) = 0

$(\cot^{2} (x) - 3) (\cot^{2} (x) - 1) = 0$

({cot}^{2} (x) - 3) ({cot}^{2} (x) - 1) = 0

$(\cot^{2} (x) - 3) (\cot^{2} (x) - 1) = 0$

Step 2

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

{cot}^{2} (x) - 3 = 0

$\cot^{2} (x) - 3 = 0$

{cot}^{2} (x) - 1 = 0

$\cot^{2} (x) - 1 = 0$

Step 3

Set

{cot}^{2} (x) - 3

$\cot^{2} (x) - 3$ equal to

0

$0$ and solve for

x

$x$ .

Tap for more steps...

Step 3.1

Set

{cot}^{2} (x) - 3

$\cot^{2} (x) - 3$ equal to

0

$0$ .

{cot}^{2} (x) - 3 = 0

$\cot^{2} (x) - 3 = 0$

Step 3.2

Solve

{cot}^{2} (x) - 3 = 0

$\cot^{2} (x) - 3 = 0$ for

x

$x$ .

Tap for more steps...

Step 3.2.1

Add

3

$3$ to both sides of the equation.

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

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

Step 3.2.2

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

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

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

Step 3.2.3

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

Tap for more steps...

Step 3.2.3.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

cot (x) = \sqrt{3}

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

Step 3.2.3.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

cot (x) = - \sqrt{3}

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

Step 3.2.3.3

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

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

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

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

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

Step 3.2.4

Set up each of the solutions to solve for

x

$x$ .

cot (x) = \sqrt{3}

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

cot (x) = - \sqrt{3}

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

Step 3.2.5

Solve for

x

$x$ in

cot (x) = \sqrt{3}

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

Tap for more steps...

Step 3.2.5.1

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 3.2.5.2

Simplify the right side.

Tap for more steps...

Step 3.2.5.2.1

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 3.2.5.3

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 3.2.5.4

Simplify

π + \frac{π}{6}

$π + \frac{π}{6}$ .

Tap for more steps...

Step 3.2.5.4.1

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 3.2.5.4.2

Combine fractions.

Tap for more steps...

Step 3.2.5.4.2.1

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 3.2.5.4.2.2

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 3.2.5.4.3

Simplify the numerator.

Tap for more steps...

Step 3.2.5.4.3.1

Move

6

$6$ to the left of

π

$π$ .

x = \frac{6 \cdot π + π}{6}

$x = \frac{6 \cdot π + π}{6}$

Step 3.2.5.4.3.2

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 3.2.5.5

Find the period of

cot (x)

$\cot (x)$ .

Tap for more steps...

Step 3.2.5.5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 3.2.5.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 3.2.5.5.3

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 3.2.5.5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 3.2.5.6

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

$n$

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 3.2.6

Solve for

x

$x$ in

cot (x) = - \sqrt{3}

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

Tap for more steps...

Step 3.2.6.1

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 3.2.6.2

Simplify the right side.

Tap for more steps...

Step 3.2.6.2.1

The exact value of

arccot (- \sqrt{3})

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

\frac{5 π}{6}

$\frac{5 π}{6}$ .

x = \frac{5 π}{6}

$x = \frac{5 π}{6}$

x = \frac{5 π}{6}

$x = \frac{5 π}{6}$

Step 3.2.6.3

The cotangent 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{5 π}{6} - π

$x = \frac{5 π}{6} - π$

Step 3.2.6.4

Simplify the expression to find the second solution.

Tap for more steps...

Step 3.2.6.4.1

Add

2 π

$2 π$ to

\frac{5 π}{6} - π

$\frac{5 π}{6} - π$ .

x = \frac{5 π}{6} - π + 2 π

$x = \frac{5 π}{6} - π + 2 π$

Step 3.2.6.4.2

The resulting angle of

\frac{11 π}{6}

$\frac{11 π}{6}$ is positive and coterminal with

\frac{5 π}{6} - π

$\frac{5 π}{6} - π$ .

x = \frac{11 π}{6}

$x = \frac{11 π}{6}$

x = \frac{11 π}{6}

$x = \frac{11 π}{6}$

Step 3.2.6.5

Find the period of

cot (x)

$\cot (x)$ .

Tap for more steps...

Step 3.2.6.5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 3.2.6.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 3.2.6.5.3

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 3.2.6.5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 3.2.6.6

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{5 π}{6} + π n, \frac{11 π}{6} + π n

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

n

$n$

x = \frac{5 π}{6} + π n, \frac{11 π}{6} + π n

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

n

$n$

Step 3.2.7

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

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

n

$n$

Step 3.2.8

Consolidate the solutions.

Tap for more steps...

Step 3.2.8.1

Consolidate

\frac{π}{6} + π n

$\frac{π}{6} + π n$ and

\frac{7 π}{6} + π n

$\frac{7 π}{6} + π n$ to

\frac{π}{6} + π n

$\frac{π}{6} + π n$ .

x = \frac{π}{6} + π n, \frac{5 π}{6} + π n, \frac{11 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

$n$

Step 3.2.8.2

Consolidate

\frac{5 π}{6} + π n

$\frac{5 π}{6} + π n$ and

\frac{11 π}{6} + π n

$\frac{11 π}{6} + π n$ to

\frac{5 π}{6} + π n

$\frac{5 π}{6} + π n$ .

x = \frac{π}{6} + π n, \frac{5 π}{6} + π n

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

n

$n$

x = \frac{π}{6} + π n, \frac{5 π}{6} + π n

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

n

$n$

x = \frac{π}{6} + π n, \frac{5 π}{6} + π n

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

n

$n$

x = \frac{π}{6} + π n, \frac{5 π}{6} + π n

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

n

$n$

Step 4

Set

{cot}^{2} (x) - 1

$\cot^{2} (x) - 1$ equal to

0

$0$ and solve for

x

$x$ .

Tap for more steps...

Step 4.1

Set

{cot}^{2} (x) - 1

$\cot^{2} (x) - 1$ equal to

0

$0$ .

{cot}^{2} (x) - 1 = 0

$\cot^{2} (x) - 1 = 0$

Step 4.2

Solve

{cot}^{2} (x) - 1 = 0

$\cot^{2} (x) - 1 = 0$ for

x

$x$ .

Tap for more steps...

Step 4.2.1

Add

1

$1$ to both sides of the equation.

{cot}^{2} (x) = 1

$\cot^{2} (x) = 1$

Step 4.2.2

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

cot (x) = \pm \sqrt{1}

$\cot (x) = \pm \sqrt{1}$

Step 4.2.3

Any root of

1

$1$ is

1

$1$ .

cot (x) = \pm 1

$\cot (x) = \pm 1$

Step 4.2.4

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

Tap for more steps...

Step 4.2.4.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

cot (x) = 1

$\cot (x) = 1$

Step 4.2.4.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

cot (x) = - 1

$\cot (x) = - 1$

Step 4.2.4.3

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

cot (x) = 1, - 1

$\cot (x) = 1, - 1$

cot (x) = 1, - 1

$\cot (x) = 1, - 1$

Step 4.2.5

Set up each of the solutions to solve for

x

$x$ .

cot (x) = 1

$\cot (x) = 1$

cot (x) = - 1

$\cot (x) = - 1$

Step 4.2.6

Solve for

x

$x$ in

cot (x) = 1

$\cot (x) = 1$ .

Tap for more steps...

Step 4.2.6.1

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

x

$x$ from inside the cotangent.

x = arccot (1)

$x = arccot (1)$

Step 4.2.6.2

Simplify the right side.

Tap for more steps...

Step 4.2.6.2.1

The exact value of

arccot (1)

$arccot (1)$ is

\frac{π}{4}

$\frac{π}{4}$ .

x = \frac{π}{4}

$x = \frac{π}{4}$

x = \frac{π}{4}

$x = \frac{π}{4}$

Step 4.2.6.3

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}{4}

$x = π + \frac{π}{4}$

Step 4.2.6.4

Simplify

π + \frac{π}{4}

$π + \frac{π}{4}$ .

Tap for more steps...

Step 4.2.6.4.1

To write

π

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

\frac{4}{4}

$\frac{4}{4}$ .

x = π \cdot \frac{4}{4} + \frac{π}{4}

$x = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 4.2.6.4.2

Combine fractions.

Tap for more steps...

Step 4.2.6.4.2.1

Combine

π

$π$ and

\frac{4}{4}

$\frac{4}{4}$ .

x = \frac{π \cdot 4}{4} + \frac{π}{4}

$x = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 4.2.6.4.2.2

Combine the numerators over the common denominator.

x = \frac{π \cdot 4 + π}{4}

$x = \frac{π \cdot 4 + π}{4}$

x = \frac{π \cdot 4 + π}{4}

$x = \frac{π \cdot 4 + π}{4}$

Step 4.2.6.4.3

Simplify the numerator.

Tap for more steps...

Step 4.2.6.4.3.1

Move

4

$4$ to the left of

π

$π$ .

x = \frac{4 \cdot π + π}{4}

$x = \frac{4 \cdot π + π}{4}$

Step 4.2.6.4.3.2

Add

4 π

$4 π$ and

π

$π$ .

x = \frac{5 π}{4}

$x = \frac{5 π}{4}$

x = \frac{5 π}{4}

$x = \frac{5 π}{4}$

x = \frac{5 π}{4}

$x = \frac{5 π}{4}$

Step 4.2.6.5

Find the period of

cot (x)

$\cot (x)$ .

Tap for more steps...

Step 4.2.6.5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 4.2.6.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 4.2.6.5.3

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 4.2.6.5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 4.2.6.6

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}{4} + π n, \frac{5 π}{4} + π n

$x = \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer

n

$n$

x = \frac{π}{4} + π n, \frac{5 π}{4} + π n

$x = \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer

n

$n$

Step 4.2.7

Solve for

x

$x$ in

cot (x) = - 1

$\cot (x) = - 1$ .

Tap for more steps...

Step 4.2.7.1

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

x

$x$ from inside the cotangent.

x = arccot (- 1)

$x = arccot (- 1)$

Step 4.2.7.2

Simplify the right side.

Tap for more steps...

Step 4.2.7.2.1

The exact value of

arccot (- 1)

$arccot (- 1)$ is

\frac{3 π}{4}

$\frac{3 π}{4}$ .

x = \frac{3 π}{4}

$x = \frac{3 π}{4}$

x = \frac{3 π}{4}

$x = \frac{3 π}{4}$

Step 4.2.7.3

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.

$x = \frac{3 π}{4} - π$

Step 4.2.7.4

Simplify the expression to find the second solution.

Tap for more steps...

Step 4.2.7.4.1

Add

$2 π$ to

$\frac{3 π}{4} - π$ .

$x = \frac{3 π}{4} - π + 2 π$

Step 4.2.7.4.2

The resulting angle of

$\frac{7 π}{4}$ is positive and coterminal with

$\frac{3 π}{4} - π$ .

$x = \frac{7 π}{4}$

Step 4.2.7.5

Find the period of

$\cot (x)$ .

Tap for more steps...

Step 4.2.7.5.1

The period of the function can be calculated using

$\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 4.2.7.5.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 4.2.7.5.3

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

$0$ and

$1$ is

$1$ .

$\frac{π}{1}$

Step 4.2.7.5.4

Divide

$π$ by

$1$ .

$π$

Step 4.2.7.6

The period of the

$\cot (x)$ function is

$π$ so values will repeat every

$π$ radians in both directions.

$x = \frac{3 π}{4} + π n, \frac{7 π}{4} + π n$ , for any integer

$n$

$x = \frac{3 π}{4} + π n, \frac{7 π}{4} + π n$ , for any integer

$n$

Step 4.2.8

List all of the solutions.

$x = \frac{π}{4} + π n, \frac{5 π}{4} + π n, \frac{3 π}{4} + π n, \frac{7 π}{4} + π n$ , for any integer

$n$

Step 4.2.9

Consolidate the answers.

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

$n$

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

$n$

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

$n$

Step 5

The final solution is all the values that make

$(\cot^{2} (x) - 3) (\cot^{2} (x) - 1) = 0$ true.

$x = \frac{π}{6} + π n, \frac{5 π}{6} + π n, \frac{π}{4} + \frac{π n}{2}$ , for any integer

$n$