Trigonometry Examples

$- \sin (2 x) > 0$

Step 1

Divide each term in $- \sin (2 x) > 0$ by $- 1$ and simplify.

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Step 1.1

Divide each term in $- \sin (2 x) > 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- \sin (2 x)}{- 1} < \frac{0}{- 1}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Dividing two negative values results in a positive value.

$\frac{\sin (2 x)}{1} < \frac{0}{- 1}$

Step 1.2.2

Divide $\sin (2 x)$ by $1$ .

$\sin (2 x) < \frac{0}{- 1}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide $0$ by $- 1$ .

$\sin (2 x) < 0$

Step 2

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

$2 x < \arcsin (0)$

Step 3

Simplify the right side.

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Step 3.1

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

$2 x < 0$

Step 4

Divide each term in $2 x < 0$ by $2$ and simplify.

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Step 4.1

Divide each term in $2 x < 0$ by $2$ .

$\frac{2 x}{2} < \frac{0}{2}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{2 x}{2} < \frac{0}{2}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x < \frac{0}{2}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $2$ .

$x < 0$

Step 5

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.

$2 x = π - 0$

Step 6

Solve for $x$ .

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Step 6.1

Simplify.

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Step 6.1.1

Multiply $- 1$ by $0$ .

$2 x = π + 0$

Step 6.1.2

Add $π$ and $0$ .

$2 x = π$

Step 6.2

Divide each term in $2 x = π$ by $2$ and simplify.

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Step 6.2.1

Divide each term in $2 x = π$ by $2$ .

$\frac{2 x}{2} = \frac{π}{2}$

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Step 6.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{2}$

Step 7

Find the period of $\sin (2 x)$ .

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Step 7.1

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

$\frac{2 π}{| b |}$

Step 7.2

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

Step 7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Step 7.4

Cancel the common factor of $2$ .

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Step 7.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 7.4.2

Divide $π$ by $1$ .

$π$

Step 8

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

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

Step 9

Consolidate the answers.

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

Step 10

Use each root to create test intervals.

$0 < x < \frac{π}{2}$

$\frac{π}{2} < x < π$

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 $0 < x < \frac{π}{2}$ to see if it makes the inequality true.

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Step 11.1.1

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

$x = 1$

Step 11.1.2

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

$- \sin (2 (1)) > 0$

Step 11.1.3

The left side $- 0.90929742$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 11.2

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

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Step 11.2.1

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

$x = 2$

Step 11.2.2

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

$- \sin (2 (2)) > 0$

Step 11.2.3

The left side $0.75680249$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 11.3

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

$0 < x < \frac{π}{2}$ False

$\frac{π}{2} < x < π$ True

$0 < x < \frac{π}{2}$ False

$\frac{π}{2} < x < π$ True

Step 12

The solution consists of all of the true intervals.

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

Step 13