Solve the Inequality for x sin(2x)>0

$\sin (2 x) > 0$

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

$2 x > \arcsin (0)$

Simplify the right side.

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

$2 x > 0$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $0$ by $2$ .

$x > 0$

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.

$2 x = π - 0$

Solve for $x$ .

Simplify.

Multiply $- 1$ by $0$ .

$2 x = π + 0$

Add $π$ and $0$ .

$2 x = π$

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

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

$\frac{2 x}{2} = \frac{π}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Divide $x$ by $1$ .

$x = \frac{π}{2}$

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

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 2 |}$

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

$\frac{2 π}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 π}{2}$

Divide $π$ by $1$ .

$π$

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 $n$

Consolidate the answers.

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

Use each root to create test intervals.

$0 < x < \frac{π}{2}$

$\frac{π}{2} < x < π$

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

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

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

$x = 1$

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

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

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

True

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

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

$x = 2$

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

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

The left side $- 0.75680249$ is not greater than the right side $0$ , which means that the given statement is false.

False

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

$0 < x < \frac{π}{2}$ True

$\frac{π}{2} < x < π$ False

$0 < x < \frac{π}{2}$ True

$\frac{π}{2} < x < π$ False

The solution consists of all of the true intervals.

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

Solve the Inequality for x sin(2x)>0

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