Precalculus Examples

\sin (x) = \frac{5}{13}

$\sin (x) = \frac{5}{13}$

Step 1

Take the inverse sine of both sides of the equation to extract

x

$x$ from inside the sine.

x = \arcsin (\frac{5}{13})

$x = \arcsin (\frac{5}{13})$

Step 2

Simplify the right side.

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Step 2.1

Evaluate

\arcsin (\frac{5}{13})

$\arcsin (\frac{5}{13})$ .

x = 0.39479111

$x = 0.39479111$

x = 0.39479111

$x = 0.39479111$

Step 3

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from

π

$π$ to find the solution in the second quadrant.

x = (3.14159265) - 0.39479111

$x = (3.14159265) - 0.39479111$

Step 4

Solve for

x

$x$ .

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Step 4.1

Remove parentheses.

x = 3.14159265 - 0.39479111

$x = 3.14159265 - 0.39479111$

Step 4.2

Remove parentheses.

x = (3.14159265) - 0.39479111

$x = (3.14159265) - 0.39479111$

Step 4.3

Subtract

0.39479111

$0.39479111$ from

3.14159265

$3.14159265$ .

x = 2.74680153

$x = 2.74680153$

x = 2.74680153

$x = 2.74680153$

Step 5

Find the period of

\sin (x)

$\sin (x)$ .

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Step 5.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 5.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 6

The period of the

\sin (x)

$\sin (x)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = 0.39479111 + 2 π n, 2.74680153 + 2 π n

$x = 0.39479111 + 2 π n, 2.74680153 + 2 π n$ , for any integer

n

$n$

\sin (x) = \frac{5}{13}

$\sin (x) = \frac{5}{13}$

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