Precalculus Examples

$\sin (x) = \frac{2}{9}$ , $0 < x < \frac{π}{2}$

Step 1

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

$0 = \frac{2}{9} - \sin (x)$

Step 2

The Intermediate Value Theorem states that, if $f$ is a real-valued continuous function on the interval $[a, b]$ , and $u$ is a number between $f (a)$ and $f (b)$ , then there is a $c$ contained in the interval $[a, b]$ such that $f (c) = u$ .

$u = f (c) = 0$

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

Calculate $f (a) = f (0) = \frac{2}{9} - \sin (0)$ .

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Step 4.1

Simplify each term.

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Step 4.1.1

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

$f (0) = \frac{2}{9} - 0$

Step 4.1.2

Multiply $- 1$ by $0$ .

$f (0) = \frac{2}{9} + 0$

Step 4.2

Add $\frac{2}{9}$ and $0$ .

$f (0) = \frac{2}{9}$

Step 5

Calculate $f (b) = f (\frac{π}{2}) = \frac{2}{9} - \sin (\frac{π}{2})$ .

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Step 5.1

Simplify each term.

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Step 5.1.1

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

$f (\frac{π}{2}) = \frac{2}{9} - 1 \cdot 1$

Step 5.1.2

Multiply $- 1$ by $1$ .

$f (\frac{π}{2}) = \frac{2}{9} - 1$

Step 5.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$f (\frac{π}{2}) = \frac{2}{9} - 1 \cdot \frac{9}{9}$

Step 5.3

Combine $- 1$ and $\frac{9}{9}$ .

$f (\frac{π}{2}) = \frac{2}{9} + \frac{- 1 \cdot 9}{9}$

Step 5.4

Combine the numerators over the common denominator.

$f (\frac{π}{2}) = \frac{2 - 1 \cdot 9}{9}$

Step 5.5

Simplify the numerator.

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Step 5.5.1

Multiply $- 1$ by $9$ .

$f (\frac{π}{2}) = \frac{2 - 9}{9}$

Step 5.5.2

Subtract $9$ from $2$ .

$f (\frac{π}{2}) = \frac{- 7}{9}$

Step 5.6

Move the negative in front of the fraction.

$f (\frac{π}{2}) = - \frac{7}{9}$

Step 6

Since $0$ is on the interval $[- \frac{7}{9}, \frac{2}{9}]$ , solve the equation for $x$ at the root.

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Step 6.1

Rewrite the equation as $\frac{2}{9} - \sin (x) = 0$ .

$\frac{2}{9} - \sin (x) = 0$

Step 6.2

Subtract $\frac{2}{9}$ from both sides of the equation.

$- \sin (x) = - \frac{2}{9}$

Step 6.3

Divide each term in $- \sin (x) = - \frac{2}{9}$ by $- 1$ and simplify.

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Step 6.3.1

Divide each term in $- \sin (x) = - \frac{2}{9}$ by $- 1$ .

$\frac{- \sin (x)}{- 1} = \frac{- \frac{2}{9}}{- 1}$

Step 6.3.2

Simplify the left side.

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Step 6.3.2.1

Dividing two negative values results in a positive value.

$\frac{\sin (x)}{1} = \frac{- \frac{2}{9}}{- 1}$

Step 6.3.2.2

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

$\sin (x) = \frac{- \frac{2}{9}}{- 1}$

Step 6.3.3

Simplify the right side.

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Step 6.3.3.1

Dividing two negative values results in a positive value.

$\sin (x) = \frac{\frac{2}{9}}{1}$

Step 6.3.3.2

Divide $\frac{2}{9}$ by $1$ .

$\sin (x) = \frac{2}{9}$

Step 6.4

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

$x = \arcsin (\frac{2}{9})$

Step 6.5

Simplify the right side.

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Step 6.5.1

Evaluate $\arcsin (\frac{2}{9})$ .

$x = 0.22409309$

Step 6.6

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.

$x = (3.14159265) - 0.22409309$

Step 6.7

Solve for $x$ .

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Step 6.7.1

Remove parentheses.

$x = 3.14159265 - 0.22409309$

Step 6.7.2

Remove parentheses.

$x = (3.14159265) - 0.22409309$

Step 6.7.3

Subtract $0.22409309$ from $3.14159265$ .

$x = 2.91749956$

Step 6.8

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

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Step 6.8.1

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

$\frac{2 π}{| b |}$

Step 6.8.2

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

$\frac{2 π}{| 1 |}$

Step 6.8.3

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

$\frac{2 π}{1}$

Step 6.8.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.9

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

$x = 0.22409309 + 2 π n, 2.91749956 + 2 π n$ , for any integer $n$

Step 7

The Intermediate Value Theorem states that there is a root $f (c) = 0$ on the interval $[- \frac{7}{9}, \frac{2}{9}]$ because $f$ is a continuous function on $[0, \frac{π}{2}]$ .

The roots on the interval $[0, \frac{π}{2}]$ are located at .

Step 8