Precalculus Examples

cot (θ) = 1

$\cot (θ) = 1$

Step 1

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

θ

$θ$ from inside the cotangent.

θ = arccot (1)

$θ = arccot (1)$

Step 2

Simplify the right side.

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Step 2.1

The exact value of

arccot (1)

$arccot (1)$ is

\frac{π}{4}

$\frac{π}{4}$ .

θ = \frac{π}{4}

$θ = \frac{π}{4}$

θ = \frac{π}{4}

$θ = \frac{π}{4}$

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.

θ = π + \frac{π}{4}

$θ = π + \frac{π}{4}$

Step 4

Simplify

π + \frac{π}{4}

$π + \frac{π}{4}$ .

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Step 4.1

To write

π

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

\frac{4}{4}

$\frac{4}{4}$ .

θ = π \cdot \frac{4}{4} + \frac{π}{4}

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

Step 4.2

Combine fractions.

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Step 4.2.1

Combine

$π$ and

$\frac{4}{4}$ .

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

Step 4.2.2

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 4 + π}{4}$

Step 4.3

Simplify the numerator.

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Step 4.3.1

Move

$4$ to the left of

$π$ .

$θ = \frac{4 \cdot π + π}{4}$

Step 4.3.2

Add

$4 π$ and

$π$ .

$θ = \frac{5 π}{4}$

Step 5

Find the period of

$\cot (θ)$ .

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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

$\cot (θ)$ function is

$π$ so values will repeat every

$π$ radians in both directions.

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

$n$

Step 7

Consolidate the answers.

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

$n$