Trigonometry Examples

2 \cot (θ) - 3 = 0

$2 \cot (θ) - 3 = 0$

Step 1

Add

3

$3$ to both sides of the equation.

2 \cot (θ) = 3

$2 \cot (θ) = 3$

Step 2

Divide each term in

2 \cot (θ) = 3

$2 \cot (θ) = 3$ by

2

$2$ and simplify.

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Step 2.1

Divide each term in

2 \cot (θ) = 3

$2 \cot (θ) = 3$ by

2

$2$ .

\frac{2 \cot (θ)}{2} = \frac{3}{2}

$\frac{2 \cot (θ)}{2} = \frac{3}{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of

2

$2$ .

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Step 2.2.1.1

Cancel the common factor.

\frac{2 \cot (θ)}{2} = \frac{3}{2}

$\frac{2 \cot (θ)}{2} = \frac{3}{2}$

Step 2.2.1.2

Divide

\cot (θ)

$\cot (θ)$ by

1

$1$ .

\cot (θ) = \frac{3}{2}

$\cot (θ) = \frac{3}{2}$

\cot (θ) = \frac{3}{2}

$\cot (θ) = \frac{3}{2}$

\cot (θ) = \frac{3}{2}

$\cot (θ) = \frac{3}{2}$

\cot (θ) = \frac{3}{2}

$\cot (θ) = \frac{3}{2}$

Step 3

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

θ

$θ$ from inside the cotangent.

θ = arccot (\frac{3}{2})

$θ = arccot (\frac{3}{2})$

Step 4

Simplify the right side.

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Step 4.1

Evaluate

arccot (\frac{3}{2})

$arccot (\frac{3}{2})$ .

θ = 33.69006752

$θ = 33.69006752$

θ = 33.69006752

$θ = 33.69006752$

Step 5

The cotangent function is positive in the first and third quadrants. To find the second solution, subtract the reference angle from

180

$180$ to find the solution in the fourth quadrant.

θ = 180 + 33.69006752

$θ = 180 + 33.69006752$

Step 6

Add

180

$180$ and

33.69006752

$33.69006752$ .

θ = 213.69006752

$θ = 213.69006752$

Step 7

Find the period of

\cot (θ)

$\cot (θ)$ .

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Step 7.1

The period of the function can be calculated using

\frac{180}{| b |}

$\frac{180}{| b |}$ .

\frac{180}{| b |}

$\frac{180}{| b |}$

Step 7.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{180}{| 1 |}

$\frac{180}{| 1 |}$

Step 7.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{180}{1}

$\frac{180}{1}$

Step 7.4

Divide

180

$180$ by

1

$1$ .

180

$180$

180

$180$

Step 8

The period of the

\cot (θ)

$\cot (θ)$ function is

180

$180$ so values will repeat every

180

$180$ degrees in both directions.

θ = 33.69006752 + 180 n, 213.69006752 + 180 n

$θ = 33.69006752 + 180 n, 213.69006752 + 180 n$ , for any integer

n

$n$

Step 9

Consolidate the answers.

θ = 33.69006752 + 180 n

$θ = 33.69006752 + 180 n$ , for any integer

n

$n$