Trigonometry Examples

$\cot (3 x) < 0$

Step 1

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

$3 x < arccot (0)$

Step 2

Simplify the right side.

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Step 2.1

The exact value of $arccot (0)$ is $\frac{π}{2}$ .

$3 x < \frac{π}{2}$

Step 3

Divide each term in $3 x < \frac{π}{2}$ by $3$ and simplify.

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Step 3.1

Divide each term in $3 x < \frac{π}{2}$ by $3$ .

$\frac{3 x}{3} < \frac{\frac{π}{2}}{3}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $3$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{3 x}{3} < \frac{\frac{π}{2}}{3}$

Step 3.2.1.2

Divide $x$ by $1$ .

$x < \frac{\frac{π}{2}}{3}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Multiply the numerator by the reciprocal of the denominator.

$x < \frac{π}{2} \cdot \frac{1}{3}$

Step 3.3.2

Multiply $\frac{π}{2} \cdot \frac{1}{3}$ .

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Step 3.3.2.1

Multiply $\frac{π}{2}$ by $\frac{1}{3}$ .

$x < \frac{π}{2 \cdot 3}$

Step 3.3.2.2

Multiply $2$ by $3$ .

$x < \frac{π}{6}$

Step 4

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

$3 x = π + \frac{π}{2}$

Step 5

Solve for $x$ .

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Step 5.1

Simplify.

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Step 5.1.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$3 x = π \cdot \frac{2}{2} + \frac{π}{2}$

Step 5.1.2

Combine $π$ and $\frac{2}{2}$ .

$3 x = \frac{π \cdot 2}{2} + \frac{π}{2}$

Step 5.1.3

Combine the numerators over the common denominator.

$3 x = \frac{π \cdot 2 + π}{2}$

Step 5.1.4

Add $π \cdot 2$ and $π$ .

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Step 5.1.4.1

Reorder $π$ and $2$ .

$3 x = \frac{2 \cdot π + π}{2}$

Step 5.1.4.2

Add $2 \cdot π$ and $π$ .

$3 x = \frac{3 \cdot π}{2}$

Step 5.2

Divide each term in $3 x = \frac{3 \cdot π}{2}$ by $3$ and simplify.

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Step 5.2.1

Divide each term in $3 x = \frac{3 \cdot π}{2}$ by $3$ .

$\frac{3 x}{3} = \frac{\frac{3 \cdot π}{2}}{3}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $3$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{\frac{3 \cdot π}{2}}{3}$

Step 5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{3 \cdot π}{2}}{3}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{3 \cdot π}{2} \cdot \frac{1}{3}$

Step 5.2.3.2

Cancel the common factor of $3$ .

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Step 5.2.3.2.1

Factor $3$ out of $3 \cdot π$ .

$x = \frac{3 (π)}{2} \cdot \frac{1}{3}$

Step 5.2.3.2.2

Cancel the common factor.

$x = \frac{3 π}{2} \cdot \frac{1}{3}$

Step 5.2.3.2.3

Rewrite the expression.

$x = \frac{π}{2}$

Step 6

Find the period of $\cot (3 x)$ .

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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 $3$ in the formula for period.

$\frac{π}{| 3 |}$

Step 6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$\frac{π}{3}$

Step 7

The period of the $\cot (3 x)$ function is $\frac{π}{3}$ so values will repeat every $\frac{π}{3}$ radians in both directions.

$x = \frac{π}{6} + \frac{π n}{3}, \frac{π}{2} + \frac{π n}{3}$ , for any integer $n$

Step 8

Consolidate the answers.

$x = \frac{π}{6} + \frac{π n}{3}$ , for any integer $n$

Step 9

Find the domain of $\cot (3 x)$ .

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Step 9.1

Set the argument in $\cot (3 x)$ equal to $π n$ to find where the expression is undefined.

$3 x = π n$ , for any integer $n$

Step 9.2

Divide each term in $3 x = π n$ by $3$ and simplify.

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Step 9.2.1

Divide each term in $3 x = π n$ by $3$ .

$\frac{3 x}{3} = \frac{π n}{3}$

Step 9.2.2

Simplify the left side.

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Step 9.2.2.1

Cancel the common factor of $3$ .

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Step 9.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{π n}{3}$

Step 9.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{π n}{3}$

Step 9.3

The domain is all values of $x$ that make the expression defined.

${x | x \neq \frac{π n}{3}}$ $n$ , for any integer $n$

Step 10

Use each root to create test intervals.

$0 < x < \frac{π}{6}$

$\frac{π}{6} < x < \frac{π}{3}$

$\frac{π}{3} < x < \frac{π}{2}$

Step 11

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 11.1

Test a value on the interval $0 < x < \frac{π}{6}$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $0 < x < \frac{π}{6}$ and see if this value makes the original inequality true.

$x = 0.26$

Step 11.1.2

Replace $x$ with $0.26$ in the original inequality.

$\cot (3 (0.26)) < 0$

Step 11.1.3

The left side $1.01085502$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 11.2

Test a value on the interval $\frac{π}{6} < x < \frac{π}{3}$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $\frac{π}{6} < x < \frac{π}{3}$ and see if this value makes the original inequality true.

$x = 1$

Step 11.2.2

Replace $x$ with $1$ in the original inequality.

$\cot (3 (1)) < 0$

Step 11.2.3

The left side $- 7.01525255$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 11.3

Test a value on the interval $\frac{π}{3} < x < \frac{π}{2}$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $\frac{π}{3} < x < \frac{π}{2}$ and see if this value makes the original inequality true.

$x = 1.31$

Step 11.3.2

Replace $x$ with $1.31$ in the original inequality.

$\cot (3 (1.31)) < 0$

Step 11.3.3

The left side $0.99399967$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 11.4

Compare the intervals to determine which ones satisfy the original inequality.

$0 < x < \frac{π}{6}$ False

$\frac{π}{6} < x < \frac{π}{3}$ True

$\frac{π}{3} < x < \frac{π}{2}$ False

$0 < x < \frac{π}{6}$ False

$\frac{π}{6} < x < \frac{π}{3}$ True

$\frac{π}{3} < x < \frac{π}{2}$ False

Step 12

The solution consists of all of the true intervals.

$\frac{π}{6} + \frac{π n}{3} < x < \frac{π}{3} + \frac{π n}{3}$ , for any integer $n$

Step 13