Trigonometry Examples

$2 | x + \frac{1}{3} | < 9$

Step 1

Divide each term in $2 | x + \frac{1}{3} | < 9$ by $2$ and simplify.

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Step 1.1

Divide each term in $2 | x + \frac{1}{3} | < 9$ by $2$ .

$\frac{2 | x + \frac{1}{3} |}{2} < \frac{9}{2}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $2$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{2 | x + \frac{1}{3} |}{2} < \frac{9}{2}$

Step 1.2.1.2

Divide $| x + \frac{1}{3} |$ by $1$ .

$| x + \frac{1}{3} | < \frac{9}{2}$

Step 2

Write $| x + \frac{1}{3} | < \frac{9}{2}$ as a piecewise.

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Step 2.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x + \frac{1}{3} \geq 0$

Step 2.2

Subtract $\frac{1}{3}$ from both sides of the inequality.

$x \geq - \frac{1}{3}$

Step 2.3

In the piece where $x + \frac{1}{3}$ is non-negative, remove the absolute value.

$x + \frac{1}{3} < \frac{9}{2}$

Step 2.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x + \frac{1}{3} < 0$

Step 2.5

Subtract $\frac{1}{3}$ from both sides of the inequality.

$x < - \frac{1}{3}$

Step 2.6

In the piece where $x + \frac{1}{3}$ is negative, remove the absolute value and multiply by $- 1$ .

$- (x + \frac{1}{3}) < \frac{9}{2}$

Step 2.7

Write as a piecewise.

${\begin{cases} x + \frac{1}{3} < \frac{9}{2} & x \geq - \frac{1}{3} \\ - (x + \frac{1}{3}) < \frac{9}{2} & x < - \frac{1}{3} \end{cases}$

Step 2.8

Apply the distributive property.

${\begin{cases} x + \frac{1}{3} < \frac{9}{2} & x \geq - \frac{1}{3} \\ - x - \frac{1}{3} < \frac{9}{2} & x < - \frac{1}{3} \end{cases}$

Step 3

Solve $x + \frac{1}{3} < \frac{9}{2}$ when $x \geq - \frac{1}{3}$ .

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Step 3.1

Move all terms not containing $x$ to the right side of the inequality.

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Step 3.1.1

Subtract $\frac{1}{3}$ from both sides of the inequality.

$x < \frac{9}{2} - \frac{1}{3}$

Step 3.1.2

To write $\frac{9}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x < \frac{9}{2} \cdot \frac{3}{3} - \frac{1}{3}$

Step 3.1.3

To write $- \frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x < \frac{9}{2} \cdot \frac{3}{3} - \frac{1}{3} \cdot \frac{2}{2}$

Step 3.1.4

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.1.4.1

Multiply $\frac{9}{2}$ by $\frac{3}{3}$ .

$x < \frac{9 \cdot 3}{2 \cdot 3} - \frac{1}{3} \cdot \frac{2}{2}$

Step 3.1.4.2

Multiply $2$ by $3$ .

$x < \frac{9 \cdot 3}{6} - \frac{1}{3} \cdot \frac{2}{2}$

Step 3.1.4.3

Multiply $\frac{1}{3}$ by $\frac{2}{2}$ .

$x < \frac{9 \cdot 3}{6} - \frac{2}{3 \cdot 2}$

Step 3.1.4.4

Multiply $3$ by $2$ .

$x < \frac{9 \cdot 3}{6} - \frac{2}{6}$

Step 3.1.5

Combine the numerators over the common denominator.

$x < \frac{9 \cdot 3 - 2}{6}$

Step 3.1.6

Simplify the numerator.

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Step 3.1.6.1

Multiply $9$ by $3$ .

$x < \frac{27 - 2}{6}$

Step 3.1.6.2

Subtract $2$ from $27$ .

$x < \frac{25}{6}$

Step 3.2

Find the intersection of $x < \frac{25}{6}$ and $x \geq - \frac{1}{3}$ .

$- \frac{1}{3} \leq x < \frac{25}{6}$

Step 4

Solve $- x - \frac{1}{3} < \frac{9}{2}$ when $x < - \frac{1}{3}$ .

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Step 4.1

Solve $- x - \frac{1}{3} < \frac{9}{2}$ for $x$ .

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Step 4.1.1

Move all terms not containing $x$ to the right side of the inequality.

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Step 4.1.1.1

Add $\frac{1}{3}$ to both sides of the inequality.

$- x < \frac{9}{2} + \frac{1}{3}$

Step 4.1.1.2

To write $\frac{9}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- x < \frac{9}{2} \cdot \frac{3}{3} + \frac{1}{3}$

Step 4.1.1.3

To write $\frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- x < \frac{9}{2} \cdot \frac{3}{3} + \frac{1}{3} \cdot \frac{2}{2}$

Step 4.1.1.4

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.1.1.4.1

Multiply $\frac{9}{2}$ by $\frac{3}{3}$ .

$- x < \frac{9 \cdot 3}{2 \cdot 3} + \frac{1}{3} \cdot \frac{2}{2}$

Step 4.1.1.4.2

Multiply $2$ by $3$ .

$- x < \frac{9 \cdot 3}{6} + \frac{1}{3} \cdot \frac{2}{2}$

Step 4.1.1.4.3

Multiply $\frac{1}{3}$ by $\frac{2}{2}$ .

$- x < \frac{9 \cdot 3}{6} + \frac{2}{3 \cdot 2}$

Step 4.1.1.4.4

Multiply $3$ by $2$ .

$- x < \frac{9 \cdot 3}{6} + \frac{2}{6}$

Step 4.1.1.5

Combine the numerators over the common denominator.

$- x < \frac{9 \cdot 3 + 2}{6}$

Step 4.1.1.6

Simplify the numerator.

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Step 4.1.1.6.1

Multiply $9$ by $3$ .

$- x < \frac{27 + 2}{6}$

Step 4.1.1.6.2

Add $27$ and $2$ .

$- x < \frac{29}{6}$

Step 4.1.2

Divide each term in $- x < \frac{29}{6}$ by $- 1$ and simplify.

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Step 4.1.2.1

Divide each term in $- x < \frac{29}{6}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} > \frac{\frac{29}{6}}{- 1}$

Step 4.1.2.2

Simplify the left side.

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Step 4.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} > \frac{\frac{29}{6}}{- 1}$

Step 4.1.2.2.2

Divide $x$ by $1$ .

$x > \frac{\frac{29}{6}}{- 1}$

Step 4.1.2.3

Simplify the right side.

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Step 4.1.2.3.1

Move the negative one from the denominator of $\frac{\frac{29}{6}}{- 1}$ .

$x > - 1 \cdot \frac{29}{6}$

Step 4.1.2.3.2

Rewrite $- 1 \cdot \frac{29}{6}$ as $- \frac{29}{6}$ .

$x > - \frac{29}{6}$

Step 4.2

Find the intersection of $x > - \frac{29}{6}$ and $x < - \frac{1}{3}$ .

$- \frac{29}{6} < x < - \frac{1}{3}$

Step 5

Find the union of the solutions.

$- \frac{29}{6} < x < \frac{25}{6}$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$- \frac{29}{6} < x < \frac{25}{6}$

Interval Notation:

$(- \frac{29}{6}, \frac{25}{6})$

Step 7