Trigonometry Examples

\sin (x) < 0

$\sin (x) < 0$ ,

\cot (x) > 0

$\cot (x) > 0$

Step 1

Simplify the first inequality.

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Step 1.1

Take the inverse sine of both sides of the equation to extract

x

$x$ from inside the sine.

x < \arcsin (0)

$x < \arcsin (0)$ and

\cot (x) > 0

$\cot (x) > 0$

Step 1.2

Simplify the right side.

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Step 1.2.1

The exact value of

\arcsin (0)

$\arcsin (0)$ is

0

$0$ .

x < 0

$x < 0$ and

\cot (x) > 0

$\cot (x) > 0$

x < 0

$x < 0$ and

\cot (x) > 0

$\cot (x) > 0$

Step 1.3

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from

π

$π$ to find the solution in the second quadrant.

x = π - 0

$x = π - 0$ and

\cot (x) > 0

$\cot (x) > 0$

Step 1.4

Subtract

0

$0$ from

π

$π$ .

x = π

$x = π$ and

\cot (x) > 0

$\cot (x) > 0$

Step 1.5

Find the period of

\sin (x)

$\sin (x)$ .

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Step 1.5.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 1.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 1.5.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 1.5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 1.6

The period of the

\sin (x)

$\sin (x)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = 2 π n, π + 2 π n

$x = 2 π n, π + 2 π n$ and

\cot (x) > 0

$\cot (x) > 0$

Step 1.7

Consolidate the answers.

x = π n

$x = π n$ and

\cot (x) > 0

$\cot (x) > 0$

Step 1.8

Use each root to create test intervals.

0 < x < π

$0 < x < π$

π < x < 2 π

$π < x < 2 π$ and

\cot (x) > 0

$\cot (x) > 0$

Step 1.9

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 1.9.1

Test a value on the interval

0 < x < π

$0 < x < π$ to see if it makes the inequality true.

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Step 1.9.1.1

Choose a value on the interval

0 < x < π

$0 < x < π$ and see if this value makes the original inequality true.

x = 2

$x = 2$ and

\cot (x) > 0

$\cot (x) > 0$

Step 1.9.1.2

Replace

x

$x$ with

2

$2$ in the original inequality.

\sin (2) < 0

$\sin (2) < 0$ and

\cot (x) > 0

$\cot (x) > 0$

Step 1.9.1.3

The left side

0.90929742

$0.90929742$ is not less than the right side

0

$0$ , which means that the given statement is false.

False and

\cot (x) > 0

$\cot (x) > 0$

False and

\cot (x) > 0

$\cot (x) > 0$

Step 1.9.2

Test a value on the interval

π < x < 2 π

$π < x < 2 π$ to see if it makes the inequality true.

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Step 1.9.2.1

Choose a value on the interval

π < x < 2 π

$π < x < 2 π$ and see if this value makes the original inequality true.

x = 5

$x = 5$ and

\cot (x) > 0

$\cot (x) > 0$

Step 1.9.2.2

Replace

x

$x$ with

5

$5$ in the original inequality.

\sin (5) < 0

$\sin (5) < 0$ and

\cot (x) > 0

$\cot (x) > 0$

Step 1.9.2.3

The left side

- 0.95892427

$- 0.95892427$ is less than the right side

0

$0$ , which means that the given statement is always true.

True and

\cot (x) > 0

$\cot (x) > 0$

True and

\cot (x) > 0

$\cot (x) > 0$

Step 1.9.3

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

0 < x < π

$0 < x < π$ False

π < x < 2 π

$π < x < 2 π$ True and

\cot (x) > 0

$\cot (x) > 0$

0 < x < π

$0 < x < π$ False

π < x < 2 π

$π < x < 2 π$ True and

\cot (x) > 0

$\cot (x) > 0$

Step 1.10

The solution consists of all of the true intervals.

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

\cot (x) > 0

$\cot (x) > 0$

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

\cot (x) > 0

$\cot (x) > 0$

Step 2

Simplify the second inequality.

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Step 2.1

Take the inverse cotangent of both sides of the equation to extract

x

$x$ from inside the cotangent.

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x > arccot (0)

$x > arccot (0)$

Step 2.2

Simplify the right side.

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Step 2.2.1

The exact value of

arccot (0)

$arccot (0)$ is

\frac{π}{2}

$\frac{π}{2}$ .

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x > \frac{π}{2}

$x > \frac{π}{2}$

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x > \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

$π$ to find the solution in the fourth quadrant.

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = π + \frac{π}{2}

$x = π + \frac{π}{2}$

Step 2.4

Simplify

π + \frac{π}{2}

$π + \frac{π}{2}$ .

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Step 2.4.1

To write

π

$π$ as a fraction with a common denominator, multiply by

\frac{2}{2}

$\frac{2}{2}$ .

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = π \cdot \frac{2}{2} + \frac{π}{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

$π$ and

\frac{2}{2}

$\frac{2}{2}$ .

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = \frac{π \cdot 2}{2} + \frac{π}{2}

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

Step 2.4.2.2

Combine the numerators over the common denominator.

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = \frac{π \cdot 2 + π}{2}

$x = \frac{π \cdot 2 + π}{2}$

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = \frac{π \cdot 2 + π}{2}

$x = \frac{π \cdot 2 + π}{2}$

Step 2.4.3

Simplify the numerator.

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Step 2.4.3.1

Move

2

$2$ to the left of

π

$π$ .

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = \frac{2 \cdot π + π}{2}

$x = \frac{2 \cdot π + π}{2}$

Step 2.4.3.2

Add

2 π

$2 π$ and

π

$π$ .

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

Step 2.5

Find the period of

\cot (x)

$\cot (x)$ .

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Step 2.5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 2.5.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{π}{1}

$\frac{π}{1}$

Step 2.5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 2.6

The period of the

\cot (x)

$\cot (x)$ function is

π

$π$ so values will repeat every

π

$π$ radians in both directions.

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = \frac{π}{2} + π n, \frac{3 π}{2} + π n

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

Step 2.7

Consolidate the answers.

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = \frac{π}{2} + π n

$x = \frac{π}{2} + π n$

Step 2.8

Use each root to create test intervals.

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

\frac{π}{2} < x < \frac{3 π}{2}

$\frac{π}{2} < x < \frac{3 π}{2}$

Step 2.9

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 2.9.1

Test a value on the interval

\frac{π}{2} < x < \frac{3 π}{2}

$\frac{π}{2} < x < \frac{3 π}{2}$ to see if it makes the inequality true.

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Step 2.9.1.1

Choose a value on the interval

\frac{π}{2} < x < \frac{3 π}{2}

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

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

x = 3

$x = 3$

Step 2.9.1.2

Replace

x

$x$ with

3

$3$ in the original inequality.

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

\cot (3) > 0

$\cot (3) > 0$

Step 2.9.1.3

The left side

- 7.01525255

$- 7.01525255$ is not greater than the right side

0

$0$ , which means that the given statement is false.

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and False

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and False

Step 2.9.2

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

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

\frac{π}{2} < x < \frac{3 π}{2}

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

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and

\frac{π}{2} < x < \frac{3 π}{2}

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

Step 2.10

Since there are no numbers that fall within the interval, this inequality has no solution.

π + 2 π n < x < 2 π + 2 π n

$π + 2 π n < x < 2 π + 2 π n$ and No solution

No solution