Precalculus Examples

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

Step 1

Subtract $1$ from both sides of the equation.

$\sin (2 x) = - 1$

Step 2

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

$2 x = \arcsin (- 1)$

Step 3

Simplify the right side.

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Step 3.1

The exact value of $\arcsin (- 1)$ is $- \frac{π}{2}$ .

$2 x = - \frac{π}{2}$

Step 4

Divide each term in $2 x = - \frac{π}{2}$ by $2$ and simplify.

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Step 4.1

Divide each term in $2 x = - \frac{π}{2}$ by $2$ .

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

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.1

Cancel the common factor.

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

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \frac{π}{2}}{2}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{π}{2} \cdot \frac{1}{2}$

Step 4.3.2

Multiply $- \frac{π}{2} \cdot \frac{1}{2}$ .

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Step 4.3.2.1

Multiply $\frac{1}{2}$ by $\frac{π}{2}$ .

$x = - \frac{π}{2 \cdot 2}$

Step 4.3.2.2

Multiply $2$ by $2$ .

$x = - \frac{π}{4}$

Step 5

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$2 x = 2 π + \frac{π}{2} + π$

Step 6

Simplify the expression to find the second solution.

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Step 6.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$2 x = 2 π + \frac{π}{2} + π - 2 π$

Step 6.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$2 x = \frac{3 π}{2}$

Step 6.3

Divide each term in $2 x = \frac{3 π}{2}$ by $2$ and simplify.

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Step 6.3.1

Divide each term in $2 x = \frac{3 π}{2}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2}$

Step 6.3.2

Simplify the left side.

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Step 6.3.2.1

Cancel the common factor of $2$ .

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Step 6.3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2}$

Step 6.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{3 π}{2}}{2}$

Step 6.3.3

Simplify the right side.

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Step 6.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{3 π}{2} \cdot \frac{1}{2}$

Step 6.3.3.2

Multiply $\frac{3 π}{2} \cdot \frac{1}{2}$ .

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Step 6.3.3.2.1

Multiply $\frac{3 π}{2}$ by $\frac{1}{2}$ .

$x = \frac{3 π}{2 \cdot 2}$

Step 6.3.3.2.2

Multiply $2$ by $2$ .

$x = \frac{3 π}{4}$

Step 7

Find the period of $\sin (2 x)$ .

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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 $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

Step 7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Step 7.4

Cancel the common factor of $2$ .

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Step 7.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 7.4.2

Divide $π$ by $1$ .

$π$

Step 8

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

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Step 8.1

Add $π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + π$

Step 8.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$π \cdot \frac{4}{4} - \frac{π}{4}$

Step 8.3

Combine fractions.

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Step 8.3.1

Combine $π$ and $\frac{4}{4}$ .

$\frac{π \cdot 4}{4} - \frac{π}{4}$

Step 8.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 4 - π}{4}$

Step 8.4

Simplify the numerator.

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Step 8.4.1

Move $4$ to the left of $π$ .

$\frac{4 \cdot π - π}{4}$

Step 8.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{4}$

Step 8.5

List the new angles.

$x = \frac{3 π}{4}$

Step 9

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

$x = \frac{3 π}{4} + π n, \frac{3 π}{4} + π n$ , for any integer $n$

Step 10

Consolidate the answers.

$x = \frac{3 π}{4} + π n$ , for any integer $n$