Trigonometry Examples

$\cot^{2} (θ) - 9 = 0$

Step 1

Add $9$ to both sides of the equation.

$\cot^{2} (θ) = 9$

Step 2

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

$\cot (θ) = \pm \sqrt{9}$

Step 3

Simplify $\pm \sqrt{9}$ .

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Step 3.1

Rewrite $9$ as $3^{2}$ .

$\cot (θ) = \pm \sqrt{3^{2}}$

Step 3.2

Pull terms out from under the radical, assuming positive real numbers.

$\cot (θ) = \pm 3$

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.

$\cot (θ) = 3$

Step 4.2

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

$\cot (θ) = - 3$

Step 4.3

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

$\cot (θ) = 3, - 3$

Step 5

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

$\cot (θ) = 3$

$\cot (θ) = - 3$

Step 6

Solve for $θ$ in $\cot (θ) = 3$ .

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Step 6.1

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

$θ = arccot (3)$

Step 6.2

Simplify the right side.

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Step 6.2.1

Evaluate $arccot (3)$ .

$θ = 18.43494882$

Step 6.3

The cotangent function is positive in the first and third quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the fourth quadrant.

$θ = 180 + 18.43494882$

Step 6.4

Add $180$ and $18.43494882$ .

$θ = 198.43494882$

Step 6.5

Find the period of $\cot (θ)$ .

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Step 6.5.1

The period of the function can be calculated using $\frac{180}{| b |}$ .

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

Step 6.5.2

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

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

Step 6.5.3

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

$\frac{180}{1}$

Step 6.5.4

Divide $180$ by $1$ .

$180$

Step 6.6

The period of the $\cot (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 18.43494882 + 180 n, 198.43494882 + 180 n$ , for any integer $n$

Step 7

Solve for $θ$ in $\cot (θ) = - 3$ .

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Step 7.1

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

$θ = arccot (- 3)$

Step 7.2

Simplify the right side.

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Step 7.2.1

Evaluate $arccot (- 3)$ .

$θ = 161.56505117$

Step 7.3

The cotangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the third quadrant.

$θ = 161.56505117 - 180$

Step 7.4

Simplify the expression to find the second solution.

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Step 7.4.1

Add $360 °$ to $161.56505117 - 180 °$ .

$θ = 161.56505117 - 180 ° + 360 °$

Step 7.4.2

The resulting angle of $341.56505117 °$ is positive and coterminal with $161.56505117 - 180$ .

$θ = 341.56505117 °$

Step 7.5

Find the period of $\cot (θ)$ .

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Step 7.5.1

The period of the function can be calculated using $\frac{180}{| b |}$ .

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

Step 7.5.2

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

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

Step 7.5.3

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

$\frac{180}{1}$

Step 7.5.4

Divide $180$ by $1$ .

$180$

Step 7.6

The period of the $\cot (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 161.56505117 + 180 n, 341.56505117 + 180 n$ , for any integer $n$

Step 8

List all of the solutions.

$θ = 18.43494882 + 180 n, 198.43494882 + 180 n, 161.56505117 + 180 n, 341.56505117 + 180 n$ , for any integer $n$

Step 9

Consolidate the solutions.

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Step 9.1

Consolidate $18.43494882 + 180 n$ and $198.43494882 + 180 n$ to $18.43494882 + 180 n$ .

$θ = 18.43494882 + 180 n, 161.56505117 + 180 n, 341.56505117 + 180 n$ , for any integer $n$

Step 9.2

Consolidate $161.56505117 + 180 n$ and $341.56505117 + 180 n$ to $161.56505117 + 180 n$ .

$θ = 18.43494882 + 180 n, 161.56505117 + 180 n$ , for any integer $n$