Trigonometry Examples

$\sin (\frac{x}{3}) < \frac{\sqrt{3}}{2}$

Step 1

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

$\frac{x}{3} < \arcsin (\frac{\sqrt{3}}{2})$

Step 2

Simplify the right side.

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Step 2.1

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

$\frac{x}{3} < \frac{π}{3}$

Step 3

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x < π$

Step 4

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.

$\frac{x}{3} = π - \frac{π}{3}$

Step 5

Solve for $x$ .

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Step 5.1

Multiply both sides of the equation by $3$ .

$3 \frac{x}{3} = 3 (π - \frac{π}{3})$

Step 5.2

Simplify both sides of the equation.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

Cancel the common factor of $3$ .

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Step 5.2.1.1.1

Cancel the common factor.

$3 \frac{x}{3} = 3 (π - \frac{π}{3})$

Step 5.2.1.1.2

Rewrite the expression.

$x = 3 (π - \frac{π}{3})$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Simplify $3 (π - \frac{π}{3})$ .

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Step 5.2.2.1.1

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

$x = 3 (π \cdot \frac{3}{3} - \frac{π}{3})$

Step 5.2.2.1.2

Simplify terms.

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Step 5.2.2.1.2.1

Combine $π$ and $\frac{3}{3}$ .

$x = 3 (\frac{π \cdot 3}{3} - \frac{π}{3})$

Step 5.2.2.1.2.2

Combine the numerators over the common denominator.

$x = 3 \frac{π \cdot 3 - π}{3}$

Step 5.2.2.1.2.3

Cancel the common factor of $3$ .

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Step 5.2.2.1.2.3.1

Cancel the common factor.

$x = 3 \frac{π \cdot 3 - π}{3}$

Step 5.2.2.1.2.3.2

Rewrite the expression.

$x = π \cdot 3 - π$

Step 5.2.2.1.3

Move $3$ to the left of $π$ .

$x = 3 π - π$

Step 5.2.2.1.4

Subtract $π$ from $3 π$ .

$x = 2 π$

Step 6

Find the period of $\sin (\frac{x}{3})$ .

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Step 6.1

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

$\frac{2 π}{| b |}$

Step 6.2

Replace $b$ with $\frac{1}{3}$ in the formula for period.

$\frac{2 π}{| \frac{1}{3} |}$

Step 6.3

$\frac{1}{3}$ is approximately $0.\overline{3}$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{3}}$

Step 6.4

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 3$

Step 6.5

Multiply $3$ by $2$ .

$6 π$

Step 7

The period of the $\sin (\frac{x}{3})$ function is $6 π$ so values will repeat every $6 π$ radians in both directions.

$x = π + 6 π n, 2 π + 6 π n$ , for any integer $n$

Step 8

Use each root to create test intervals.

$π < x < 2 π$

$2 π < x < 7 π$

$7 π < x < 8 π$

Step 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 9.1

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

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Step 9.1.1

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

$x = 5$

Step 9.1.2

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

$\sin (\frac{5}{3}) < \frac{\sqrt{3}}{2}$

Step 9.1.3

The left side $0.99540795$ is not less than the right side $0.8660254$ , which means that the given statement is false.

False

Step 9.2

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

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Step 9.2.1

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

$x = 9$

Step 9.2.2

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

$\sin (\frac{9}{3}) < \frac{\sqrt{3}}{2}$

Step 9.2.3

The left side $0.14112$ is less than the right side $0.8660254$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $7 π < x < 8 π$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $7 π < x < 8 π$ and see if this value makes the original inequality true.

$x = 24$

Step 9.3.2

Replace $x$ with $24$ in the original inequality.

$\sin (\frac{24}{3}) < \frac{\sqrt{3}}{2}$

Step 9.3.3

The left side $0.98935824$ is not less than the right side $0.8660254$ , which means that the given statement is false.

False

Step 9.4

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

$π < x < 2 π$ False

$2 π < x < 7 π$ True

$7 π < x < 8 π$ False

$π < x < 2 π$ False

$2 π < x < 7 π$ True

$7 π < x < 8 π$ False

Step 10

The solution consists of all of the true intervals.

$2 π + 6 π n < x < 7 π + 6 π n$ , for any integer $n$

Step 11

Convert the inequality to interval notation.

$(2 π + 6 π n, 7 π + 6 π n)$

Step 12