Trigonometry Examples

$\sec (x) > 0$ , $\cot (x) > 0$

Step 1

The range of secant is $y \leq - 1$ and $y \geq 1$ . Since $0$ does not fall in this range, there is no solution.

No solution

Step 2

Solve $\cot (x) > 0$ for $x$ .

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Step 2.1

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

$x > arccot (0)$

Step 2.2

Simplify the right side.

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Step 2.2.1

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

$x > \frac{π}{2}$

Step 2.3

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.

$x = π + \frac{π}{2}$

Step 2.4

Simplify $π + \frac{π}{2}$ .

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Step 2.4.1

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

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

Step 2.4.2

Combine fractions.

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Step 2.4.2.1

Combine $π$ and $\frac{2}{2}$ .

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

Step 2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 + π}{2}$

Step 2.4.3

Simplify the numerator.

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Step 2.4.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π + π}{2}$

Step 2.4.3.2

Add $2 π$ and $π$ .

$x = \frac{3 π}{2}$

Step 2.5

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

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Step 2.5.1

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

$\frac{π}{| b |}$

Step 2.5.2

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

$\frac{π}{| 1 |}$

Step 2.5.3

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

$\frac{π}{1}$

Step 2.5.4

Divide $π$ by $1$ .

$π$

Step 2.6

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

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

Step 2.7

Consolidate the answers.

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

Step 2.8

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

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Step 2.8.1

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

$x = π n$ , for any integer $n$

Step 2.8.2

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

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

Step 2.9

Use each root to create test intervals.

$0 < x < \frac{π}{2}$

$\frac{π}{2} < x < π$

$π < x < \frac{3 π}{2}$

Step 2.10

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 2.10.1

Test a value on the interval $0 < x < \frac{π}{2}$ to see if it makes the inequality true.

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Step 2.10.1.1

Choose a value on the interval $0 < x < \frac{π}{2}$ and see if this value makes the original inequality true.

$x = 1$

Step 2.10.1.2

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

$\cot (1) > 0$

Step 2.10.1.3

The left side $0.64209261$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.10.2

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

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Step 2.10.2.1

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

$x = 2$

Step 2.10.2.2

Replace $x$ with $2$ in the original inequality.

$\cot (2) > 0$

Step 2.10.2.3

The left side $- 0.45765755$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 2.10.3

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

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Step 2.10.3.1

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

$x = 4$

Step 2.10.3.2

Replace $x$ with $4$ in the original inequality.

$\cot (4) > 0$

Step 2.10.3.3

The left side $0.86369115$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.10.4

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

$0 < x < \frac{π}{2}$ True

$\frac{π}{2} < x < π$ False

$π < x < \frac{3 π}{2}$ True

$0 < x < \frac{π}{2}$ True

$\frac{π}{2} < x < π$ False

$π < x < \frac{3 π}{2}$ True

Step 2.11

The solution consists of all of the true intervals.

$0 + π n < x < \frac{π}{2} + π n$ or $π + π n < x < \frac{3 π}{2} + π n$ , for any integer $n$

Step 2.12

Combine the intervals.

$π n < x < \frac{π}{2} + π n or π + π n < x < \frac{3 π}{2} + π n$ , for any integer $n$

Step 3

Find the intersection of No solution and $π n < x < \frac{π}{2} + π n or π + π n < x < \frac{3 π}{2} + π n$ .

No solution