Trigonometry Examples

cot (x) = - 1

$\cot (x) = - 1$

Step 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 2

Simplify the right side.

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Step 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 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{3 π}{4} - π

$x = \frac{3 π}{4} - π$

Step 4

Simplify the expression to find the second solution.

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Step 4.1

Add

2 π

$2 π$ to

\frac{3 π}{4} - π

$\frac{3 π}{4} - π$ .

x = \frac{3 π}{4} - π + 2 π

$x = \frac{3 π}{4} - π + 2 π$

Step 4.2

The resulting angle of

\frac{7 π}{4}

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

\frac{3 π}{4} - π

$\frac{3 π}{4} - π$ .

x = \frac{7 π}{4}

$x = \frac{7 π}{4}$

x = \frac{7 π}{4}

$x = \frac{7 π}{4}$

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{3 π}{4} + π n, \frac{7 π}{4} + π n

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

n

$n$

Step 7

Consolidate the answers.

x = \frac{3 π}{4} + π n

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

n

$n$