Trigonometry Examples

$\cot (θ) + \sqrt{3} = 0$

Step 1

Subtract $\sqrt{3}$ from both sides of the equation.

$\cot (θ) = - \sqrt{3}$

Step 2

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

$θ = arccot (- \sqrt{3})$

Step 3

Simplify the right side.

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Step 3.1

The exact value of $arccot (- \sqrt{3})$ is $\frac{5 π}{6}$ .

$θ = \frac{5 π}{6}$

Step 4

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.

$θ = \frac{5 π}{6} - π$

Step 5

Simplify the expression to find the second solution.

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Step 5.1

Add $2 π$ to $\frac{5 π}{6} - π$ .

$θ = \frac{5 π}{6} - π + 2 π$

Step 5.2

The resulting angle of $\frac{11 π}{6}$ is positive and coterminal with $\frac{5 π}{6} - π$ .

$θ = \frac{11 π}{6}$

Step 6

Find the period of $\cot (θ)$ .

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Step 6.1

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

$\frac{π}{| b |}$

Step 6.2

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

$\frac{π}{| 1 |}$

Step 6.3

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

$\frac{π}{1}$

Step 6.4

Divide $π$ by $1$ .

$π$

Step 7

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

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

Step 8

Consolidate the answers.

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