Trigonometry Examples

$\sin (x) > 0$ , $\tan (x) < 0$

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 $x$ from inside the sine.

$x > \arcsin (0)$ and $\tan (x) < 0$

Step 1.2

Simplify the right side.

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Step 1.2.1

The exact value of $\arcsin (0)$ is $0$ .

$x > 0$ and $\tan (x) < 0$

Step 1.3

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

$x = π - 0$ and $\tan (x) < 0$

Step 1.4

Subtract $0$ from $π$ .

$x = π$ and $\tan (x) < 0$

Step 1.5

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

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Step 1.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 1.5.2

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

$\frac{2 π}{| 1 |}$

Step 1.5.3

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

$\frac{2 π}{1}$

Step 1.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.6

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 2 π n, π + 2 π n$ and $\tan (x) < 0$

Step 1.7

Consolidate the answers.

$x = π n$ and $\tan (x) < 0$

Step 1.8

Use each root to create test intervals.

$0 < x < π$

$π < x < 2 π$ and $\tan (x) < 0$

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

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Step 1.9.1.1

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

$x = 2$ and $\tan (x) < 0$

Step 1.9.1.2

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

$\sin (2) > 0$ and $\tan (x) < 0$

Step 1.9.1.3

The left side $0.90929742$ is greater than the right side $0$ , which means that the given statement is always true.

True and $\tan (x) < 0$

Step 1.9.2

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

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Step 1.9.2.1

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

$x = 5$ and $\tan (x) < 0$

Step 1.9.2.2

Replace $x$ with $5$ in the original inequality.

$\sin (5) > 0$ and $\tan (x) < 0$

Step 1.9.2.3

The left side $- 0.95892427$ is not greater than the right side $0$ , which means that the given statement is false.

False and $\tan (x) < 0$

Step 1.9.3

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

$0 < x < π$ True

$π < x < 2 π$ False and $\tan (x) < 0$

$0 < x < π$ True

$π < x < 2 π$ False and $\tan (x) < 0$

Step 1.10

The solution consists of all of the true intervals.

$0 + 2 π n < x < π + 2 π n$ and $\tan (x) < 0$

Step 2

Simplify the second inequality.

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Step 2.1

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

$0 + 2 π n < x < π + 2 π n$ and $x < \arctan (0)$

Step 2.2

Simplify the right side.

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Step 2.2.1

The exact value of $\arctan (0)$ is $0$ .

$0 + 2 π n < x < π + 2 π n$ and $x < 0$

Step 2.3

The tangent 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.

$0 + 2 π n < x < π + 2 π n$ and $x = π + 0$

Step 2.4

Add $π$ and $0$ .

$0 + 2 π n < x < π + 2 π n$ and $x = π$

Step 2.5

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

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

$0 + 2 π n < x < π + 2 π n$ and $x = π n, π + π n$

Step 2.7

Consolidate the answers.

$0 + 2 π n < x < π + 2 π n$ and $x = π n$

Step 2.8

Use each root to create test intervals.

$0 + 2 π n < x < π + 2 π n$ and $0 < x < π$

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

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Step 2.9.1.1

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

$0 + 2 π n < x < π + 2 π n$ and $x = 2$

Step 2.9.1.2

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

$0 + 2 π n < x < π + 2 π n$ and $\tan (2) < 0$

Step 2.9.1.3

The left side $- 2.18503986$ is less than the right side $0$ , which means that the given statement is always true.

$0 + 2 π n < x < π + 2 π n$ and True

Step 2.9.2

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

$0 + 2 π n < x < π + 2 π n$ and $0 < x < π$ True

Step 2.10

The solution consists of all of the true intervals.

$0 + 2 π n < x < π + 2 π n$ and $0 + π n < x < π + π n$

Step 3