Precalculus Examples

\ln (\sin (x)) = 0

$\ln (\sin (x)) = 0$

Step 1

To solve for

x

$x$ , rewrite the equation using properties of logarithms.

e^{\ln (\sin (x))} = e^{0}

$e^{\ln (\sin (x))} = e^{0}$

Step 2

Rewrite

\ln (\sin (x)) = 0

$\ln (\sin (x)) = 0$ in exponential form using the definition of a logarithm. If

x

$x$ and

b

$b$ are positive real numbers and

b \neq 1

$b \neq 1$ , then

\log_{b} (x) = y

$\log_{b} (x) = y$ is equivalent to

b^{y} = x

$b^{y} = x$ .

e^{0} = \sin (x)

$e^{0} = \sin (x)$

Step 3

Solve for

x

$x$ .

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Step 3.1

Rewrite the equation as

\sin (x) = e^{0}

$\sin (x) = e^{0}$ .

\sin (x) = e^{0}

$\sin (x) = e^{0}$

Step 3.2

Anything raised to

0

$0$ is

1

$1$ .

\sin (x) = 1

$\sin (x) = 1$

Step 3.3

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 3.4

Simplify the right side.

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Step 3.4.1

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 3.5

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 = π - \frac{π}{2}

$x = π - \frac{π}{2}$

Step 3.6

Simplify

π - \frac{π}{2}

$π - \frac{π}{2}$ .

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Step 3.6.1

To write

π

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

\frac{2}{2}

$\frac{2}{2}$ .

x = π \cdot \frac{2}{2} - \frac{π}{2}

$x = π \cdot \frac{2}{2} - \frac{π}{2}$

Step 3.6.2

Combine fractions.

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Step 3.6.2.1

Combine

π

$π$ and

\frac{2}{2}

$\frac{2}{2}$ .

x = \frac{π \cdot 2}{2} - \frac{π}{2}

$x = \frac{π \cdot 2}{2} - \frac{π}{2}$

Step 3.6.2.2

Combine the numerators over the common denominator.

x = \frac{π \cdot 2 - π}{2}

$x = \frac{π \cdot 2 - π}{2}$

x = \frac{π \cdot 2 - π}{2}

$x = \frac{π \cdot 2 - π}{2}$

Step 3.6.3

Simplify the numerator.

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Step 3.6.3.1

Move

2

$2$ to the left of

π

$π$ .

x = \frac{2 \cdot π - π}{2}

$x = \frac{2 \cdot π - π}{2}$

Step 3.6.3.2

Subtract

π

$π$ from

2 π

$2 π$ .

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

Step 3.7

Find the period of

\sin (x)

$\sin (x)$ .

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Step 3.7.1

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 3.7.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 3.7.3

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 3.7.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 3.8

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{π}{2} + 2 π n

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

n

$n$

x = \frac{π}{2} + 2 π n

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

n

$n$