Trigonometry Examples

$\cot (x) = - \sqrt{15}$

Step 1

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

$x = arccot (- \sqrt{15})$

Step 2

Simplify the right side.

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Step 2.1

Evaluate $arccot (- \sqrt{15})$ .

$x = 2.88891239$

Step 3

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.

$x = 2.88891239 - (3.14159265)$

Step 4

Simplify the expression to find the second solution.

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Step 4.1

Add $2 π$ to $2.88891239 - (3.14159265)$ .

$x = 2.88891239 - (3.14159265) + 2 π$

Step 4.2

The resulting angle of $6.03050505$ is positive and coterminal with $2.88891239 - (3.14159265)$ .

$x = 6.03050505$

Step 5

Find the period of $\cot (x)$ .

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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 $\cot (x)$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = 2.88891239 + π n, 6.03050505 + π n$ , for any integer $n$

Step 7

Consolidate $2.88891239 + π n$ and $6.03050505 + π n$ to $2.88891239 + π n$ .

$x = 2.88891239 + π n$ , for any integer $n$