Precalculus Examples

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

Step 1

Divide each term in $2 \sin (x - 1) = 0$ by $2$ and simplify.

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Step 1.1

Divide each term in $2 \sin (x - 1) = 0$ by $2$ .

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

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $2$ .

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Step 1.2.1.1

Cancel the common factor.

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

Step 1.2.1.2

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

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

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide $0$ by $2$ .

$\sin (x - 1) = 0$

Step 2

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

$x - 1 = \arcsin (0)$

Step 3

Simplify the right side.

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Step 3.1

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

$x - 1 = 0$

Step 4

Add $1$ to both sides of the equation.

$x = 1$

Step 5

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

$x - 1 = π - 0$

Step 6

Solve for $x$ .

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Step 6.1

Subtract $0$ from $π$ .

$x - 1 = π$

Step 6.2

Add $1$ to both sides of the equation.

$x = π + 1$

Step 7

Find the period of $\sin (x - 1)$ .

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Step 7.1

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

$\frac{2 π}{| b |}$

Step 7.2

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

$\frac{2 π}{| 1 |}$

Step 7.3

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

$\frac{2 π}{1}$

Step 7.4

Divide $2 π$ by $1$ .

$2 π$

Step 8

The period of the $\sin (x - 1)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

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

Step 9

Consolidate $1 + 2 π n$ and $π + 1 + 2 π n$ to $1 + π n$ .

$x = 1 + π n$ , for any integer $n$