Calculus Examples

$\sin (2 x) \geq \sin (x)$

Step 1

Subtract $\sin (x)$ from both sides of the inequality.

$\sin (2 x) - \sin (x) \geq 0$

Step 2

Apply the sine double-angle identity.

$2 \sin (x) \cos (x) - \sin (x) \geq 0$

Step 3

Factor $\sin (x)$ out of $2 \sin (x) \cos (x) - \sin (x)$ .

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Step 3.1

Factor $\sin (x)$ out of $2 \sin (x) \cos (x)$ .

$\sin (x) (2 \cos (x)) - \sin (x) = 0$

Step 3.2

Factor $\sin (x)$ out of $- \sin (x)$ .

$\sin (x) (2 \cos (x)) + \sin (x) \cdot - 1 = 0$

Step 3.3

Factor $\sin (x)$ out of $\sin (x) (2 \cos (x)) + \sin (x) \cdot - 1$ .

$\sin (x) (2 \cos (x) - 1) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\sin (x) = 0$

$2 \cos (x) - 1 = 0$

Step 5

Set $\sin (x)$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $\sin (x)$ equal to $0$ .

$\sin (x) = 0$

Step 5.2

Solve $\sin (x) = 0$ for $x$ .

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Step 5.2.1

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

$x = \arcsin (0)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

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

$x = 0$

Step 5.2.3

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.

$x = π - 0$

Step 5.2.4

Subtract $0$ from $π$ .

$x = π$

Step 5.2.5

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

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Step 5.2.5.1

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

$\frac{2 π}{| b |}$

Step 5.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 5.2.5.3

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

$\frac{2 π}{1}$

Step 5.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.6

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

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 6

Set $2 \cos (x) - 1$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $2 \cos (x) - 1$ equal to $0$ .

$2 \cos (x) - 1 = 0$

Step 6.2

Solve $2 \cos (x) - 1 = 0$ for $x$ .

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Step 6.2.1

Add $1$ to both sides of the equation.

$2 \cos (x) = 1$

Step 6.2.2

Divide each term in $2 \cos (x) = 1$ by $2$ and simplify.

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Step 6.2.2.1

Divide each term in $2 \cos (x) = 1$ by $2$ .

$\frac{2 \cos (x)}{2} = \frac{1}{2}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{2 \cos (x)}{2} = \frac{1}{2}$

Step 6.2.2.2.1.2

Divide $\cos (x)$ by $1$ .

$\cos (x) = \frac{1}{2}$

Step 6.2.3

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

$x = \arccos (\frac{1}{2})$

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$x = \frac{π}{3}$

Step 6.2.5

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - \frac{π}{3}$

Step 6.2.6

Simplify $2 π - \frac{π}{3}$ .

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Step 6.2.6.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = 2 π \cdot \frac{3}{3} - \frac{π}{3}$

Step 6.2.6.2

Combine fractions.

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Step 6.2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

$x = \frac{2 π \cdot 3}{3} - \frac{π}{3}$

Step 6.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 6.2.6.3

Simplify the numerator.

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Step 6.2.6.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 6.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 6.2.7

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

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Step 6.2.7.1

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

$\frac{2 π}{| b |}$

Step 6.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 6.2.7.3

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

$\frac{2 π}{1}$

Step 6.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.8

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

$x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 7

The final solution is all the values that make $\sin (x) (2 \cos (x) - 1) = 0$ true.

$x = 2 π n, π + 2 π n, \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 8

Consolidate $2 π n$ and $π + 2 π n$ to $π n$ .

$x = π n, \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 9

Use each root to create test intervals.

$0 < x < \frac{π}{3}$

$\frac{π}{3} < x < π$

$π < x < \frac{5 π}{3}$

$\frac{5 π}{3} < x < 2 π$

$2 π < x < \frac{7 π}{3}$

$\frac{7 π}{3} < x < \frac{11 π}{3}$

Step 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 10.1

Test a value on the interval $0 < x < \frac{π}{3}$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $0 < x < \frac{π}{3}$ and see if this value makes the original inequality true.

$x = 1$

Step 10.1.2

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

$\sin (2 (1)) \geq \sin (1)$

Step 10.1.3

The left side $0.90929742$ is greater than the right side $0.84147098$ , which means that the given statement is always true.

True

Step 10.2

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

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Step 10.2.1

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

$x = 2$

Step 10.2.2

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

$\sin (2 (2)) \geq \sin (2)$

Step 10.2.3

The left side $- 0.75680249$ is less than the right side $0.90929742$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $π < x < \frac{5 π}{3}$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $π < x < \frac{5 π}{3}$ and see if this value makes the original inequality true.

$x = 4$

Step 10.3.2

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

$\sin (2 (4)) \geq \sin (4)$

Step 10.3.3

The left side $0.98935824$ is greater than the right side $- 0.75680249$ , which means that the given statement is always true.

True

Step 10.4

Test a value on the interval $\frac{5 π}{3} < x < 2 π$ to see if it makes the inequality true.

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Step 10.4.1

Choose a value on the interval $\frac{5 π}{3} < x < 2 π$ and see if this value makes the original inequality true.

$x = 6$

Step 10.4.2

Replace $x$ with $6$ in the original inequality.

$\sin (2 (6)) \geq \sin (6)$

Step 10.4.3

The left side $- 0.53657291$ is less than the right side $- 0.27941549$ , which means that the given statement is false.

False

Step 10.5

Test a value on the interval $2 π < x < \frac{7 π}{3}$ to see if it makes the inequality true.

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Step 10.5.1

Choose a value on the interval $2 π < x < \frac{7 π}{3}$ and see if this value makes the original inequality true.

$x = 7$

Step 10.5.2

Replace $x$ with $7$ in the original inequality.

$\sin (2 (7)) \geq \sin (7)$

Step 10.5.3

The left side $0.99060735$ is greater than the right side $0.65698659$ , which means that the given statement is always true.

True

Step 10.6

Test a value on the interval $\frac{7 π}{3} < x < \frac{11 π}{3}$ to see if it makes the inequality true.

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Step 10.6.1

Choose a value on the interval $\frac{7 π}{3} < x < \frac{11 π}{3}$ and see if this value makes the original inequality true.

$x = 9$

Step 10.6.2

Replace $x$ with $9$ in the original inequality.

$\sin (2 (9)) \geq \sin (9)$

Step 10.6.3

The left side $- 0.75098724$ is less than the right side $0.41211848$ , which means that the given statement is false.

False

Step 10.7

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

$0 < x < \frac{π}{3}$ True

$\frac{π}{3} < x < π$ False

$π < x < \frac{5 π}{3}$ True

$\frac{5 π}{3} < x < 2 π$ False

$2 π < x < \frac{7 π}{3}$ True

$\frac{7 π}{3} < x < \frac{11 π}{3}$ False

$0 < x < \frac{π}{3}$ True

$\frac{π}{3} < x < π$ False

$π < x < \frac{5 π}{3}$ True

$\frac{5 π}{3} < x < 2 π$ False

$2 π < x < \frac{7 π}{3}$ True

$\frac{7 π}{3} < x < \frac{11 π}{3}$ False

Step 11

The solution consists of all of the true intervals.

$0 + 2 π n \leq x \leq \frac{π}{3} + 2 π n$ or $π + 2 π n \leq x \leq \frac{5 π}{3} + 2 π n$ or $2 π + 2 π n \leq x \leq \frac{7 π}{3} + 2 π n$ , for any integer $n$

Step 12

Combine the intervals.

$π + 2 π n \leq x \leq \frac{5 π}{3} + 2 π n or 2 π n \leq x \leq \frac{π}{3} + 2 π n or 2 π + 2 π n \leq x \leq \frac{7 π}{3} + 2 π n$ , for any integer $n$

Step 13