Precalculus Examples

2 \sin (x - 1) = 0

$2 \sin (x - 1) = 0$

Step 1

Divide each term in

2 \sin (x - 1) = 0

$2 \sin (x - 1) = 0$ by

2

$2$ and simplify.

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Step 1.1

Divide each term in

2 \sin (x - 1) = 0

$2 \sin (x - 1) = 0$ by

2

$2$ .

\frac{2 \sin (x - 1)}{2} = \frac{0}{2}

$\frac{2 \sin (x - 1)}{2} = \frac{0}{2}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of

2

$2$ .

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Step 1.2.1.1

Cancel the common factor.

\frac{2 \sin (x - 1)}{2} = \frac{0}{2}

$\frac{2 \sin (x - 1)}{2} = \frac{0}{2}$

Step 1.2.1.2

Divide

\sin (x - 1)

$\sin (x - 1)$ by

1

$1$ .

\sin (x - 1) = \frac{0}{2}

$\sin (x - 1) = \frac{0}{2}$

\sin (x - 1) = \frac{0}{2}

$\sin (x - 1) = \frac{0}{2}$

\sin (x - 1) = \frac{0}{2}

$\sin (x - 1) = \frac{0}{2}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide

0

$0$ by

2

$2$ .

\sin (x - 1) = 0

$\sin (x - 1) = 0$

\sin (x - 1) = 0

$\sin (x - 1) = 0$

\sin (x - 1) = 0

$\sin (x - 1) = 0$

Step 2

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

x

$x$ from inside the sine.

x - 1 = \arcsin (0)

$x - 1 = \arcsin (0)$

Step 3

Simplify the right side.

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Step 3.1

The exact value of

\arcsin (0)

$\arcsin (0)$ is

0

$0$ .

x - 1 = 0

$x - 1 = 0$

x - 1 = 0

$x - 1 = 0$

Step 4

Add

1

$1$ to both sides of the equation.

x = 1

$x = 1$

Step 5

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 - 1 = π - 0

$x - 1 = π - 0$

Step 6

Solve for

x

$x$ .

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Step 6.1

Subtract

0

$0$ from

π

$π$ .

x - 1 = π

$x - 1 = π$

Step 6.2

Add

1

$1$ to both sides of the equation.

x = π + 1

$x = π + 1$

x = π + 1

$x = π + 1$

Step 7

Find the period of

\sin (x - 1)

$\sin (x - 1)$ .

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Step 7.1

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 7.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 7.3

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 7.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 8

The period of the

\sin (x - 1)

$\sin (x - 1)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = 1 + 2 π n, π + 1 + 2 π n

$x = 1 + 2 π n, π + 1 + 2 π n$ , for any integer

n

$n$

Step 9

Consolidate

1 + 2 π n

$1 + 2 π n$ and

π + 1 + 2 π n

$π + 1 + 2 π n$ to

1 + π n

$1 + π n$ .

x = 1 + π n

$x = 1 + π n$ , for any integer

n

$n$