Precalculus Examples

sin (x) (sin (x) + 1) = 0

$\sin (x) (\sin (x) + 1) = 0$

Step 1

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

sin (x) = 0

$\sin (x) = 0$

sin (x) + 1 = 0

$\sin (x) + 1 = 0$

Step 2

Set

sin (x)

$\sin (x)$ equal to

0

$0$ and solve for

x

$x$ .

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Step 2.1

Set

sin (x)

$\sin (x)$ equal to

0

$0$ .

sin (x) = 0

$\sin (x) = 0$

Step 2.2

Solve

sin (x) = 0

$\sin (x) = 0$ for

x

$x$ .

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Step 2.2.1

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

x

$x$ from inside the sine.

x = arcsin (0)

$x = \arcsin (0)$

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

The exact value of

arcsin (0)

$\arcsin (0)$ is

0

$0$ .

x = 0

$x = 0$

x = 0

$x = 0$

Step 2.2.3

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 = π - 0

$x = π - 0$

Step 2.2.4

Subtract

0

$0$ from

π

$π$ .

x = π

$x = π$

Step 2.2.5

Find the period of

sin (x)

$\sin (x)$ .

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Step 2.2.5.1

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 2.2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 2.2.5.3

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 2.2.5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 2.2.6

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 = 2 π n, π + 2 π n

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

n

$n$

x = 2 π n, π + 2 π n

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

n

$n$

x = 2 π n, π + 2 π n

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

n

$n$

Step 3

Set

sin (x) + 1

$\sin (x) + 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.1

Set

sin (x) + 1

$\sin (x) + 1$ equal to

0

$0$ .

sin (x) + 1 = 0

$\sin (x) + 1 = 0$

Step 3.2

Solve

sin (x) + 1 = 0

$\sin (x) + 1 = 0$ for

x

$x$ .

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Step 3.2.1

Subtract

1

$1$ from both sides of the equation.

sin (x) = - 1

$\sin (x) = - 1$

Step 3.2.2

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

x

$x$ from inside the sine.

x = arcsin (- 1)

$x = \arcsin (- 1)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

The exact value of

arcsin (- 1)

$\arcsin (- 1)$ is

- \frac{π}{2}

$- \frac{π}{2}$ .

x = - \frac{π}{2}

$x = - \frac{π}{2}$

x = - \frac{π}{2}

$x = - \frac{π}{2}$

Step 3.2.4

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from

2 π

$2 π$ , to find a reference angle. Next, add this reference angle to

π

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

x = 2 π + \frac{π}{2} + π

$x = 2 π + \frac{π}{2} + π$

Step 3.2.5

Simplify the expression to find the second solution.

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Step 3.2.5.1

Subtract

2 π

$2 π$ from

2 π + \frac{π}{2} + π

$2 π + \frac{π}{2} + π$ .

x = 2 π + \frac{π}{2} + π - 2 π

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 3.2.5.2

The resulting angle of

\frac{3 π}{2}

$\frac{3 π}{2}$ is positive, less than

2 π

$2 π$ , and coterminal with

2 π + \frac{π}{2} + π

$2 π + \frac{π}{2} + π$ .

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

Step 3.2.6

Find the period of

sin (x)

$\sin (x)$ .

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Step 3.2.6.1

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 3.2.6.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 3.2.6.3

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 3.2.6.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 3.2.7

Add

2 π

$2 π$ to every negative angle to get positive angles.

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Step 3.2.7.1

Add

2 π

$2 π$ to

- \frac{π}{2}

$- \frac{π}{2}$ to find the positive angle.

- \frac{π}{2} + 2 π

$- \frac{π}{2} + 2 π$

Step 3.2.7.2

To write

2 π

$2 π$ as a fraction with a common denominator, multiply by

\frac{2}{2}

$\frac{2}{2}$ .

2 π \cdot \frac{2}{2} - \frac{π}{2}

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 3.2.7.3

Combine fractions.

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Step 3.2.7.3.1

Combine

2 π

$2 π$ and

\frac{2}{2}

$\frac{2}{2}$ .

\frac{2 π \cdot 2}{2} - \frac{π}{2}

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 3.2.7.3.2

Combine the numerators over the common denominator.

\frac{2 π \cdot 2 - π}{2}

$\frac{2 π \cdot 2 - π}{2}$

\frac{2 π \cdot 2 - π}{2}

$\frac{2 π \cdot 2 - π}{2}$

Step 3.2.7.4

Simplify the numerator.

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Step 3.2.7.4.1

Multiply

2

$2$ by

2

$2$ .

\frac{4 π - π}{2}

$\frac{4 π - π}{2}$

Step 3.2.7.4.2

Subtract

π

$π$ from

4 π

$4 π$ .

\frac{3 π}{2}

$\frac{3 π}{2}$

\frac{3 π}{2}

$\frac{3 π}{2}$

Step 3.2.7.5

List the new angles.

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

Step 3.2.8

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 = \frac{3 π}{2} + 2 π n, \frac{3 π}{2} + 2 π n

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

n

$n$

x = \frac{3 π}{2} + 2 π n, \frac{3 π}{2} + 2 π n

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

n

$n$

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

$n$

Step 4

The final solution is all the values that make

$\sin (x) (\sin (x) + 1) = 0$ true.

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

$n$

Step 5

Consolidate

$2 π n$ and

$π + 2 π n$ to

$π n$ .

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

$n$