Precalculus Examples

$- \sin (x) = \sin (x) + \sqrt{2}$

Step 1

Move all terms containing $\sin (x)$ to the left side of the equation.

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Step 1.1

Subtract $\sin (x)$ from both sides of the equation.

$- \sin (x) - \sin (x) = \sqrt{2}$

Step 1.2

Subtract $\sin (x)$ from $- \sin (x)$ .

$- 2 \sin (x) = \sqrt{2}$

Step 2

Divide each term in $- 2 \sin (x) = \sqrt{2}$ by $- 2$ and simplify.

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Step 2.1

Divide each term in $- 2 \sin (x) = \sqrt{2}$ by $- 2$ .

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $- 2$ .

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Step 2.2.1.1

Cancel the common factor.

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

Step 2.2.1.2

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

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

Step 2.3

Simplify the right side.

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Step 2.3.1

Move the negative in front of the fraction.

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

Step 3

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

$x = \arcsin (- \frac{\sqrt{2}}{2})$

Step 4

Simplify the right side.

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Step 4.1

The exact value of $\arcsin (- \frac{\sqrt{2}}{2})$ is $- \frac{π}{4}$ .

$x = - \frac{π}{4}$

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.

$x = 2 π + \frac{π}{4} + π$

Step 6

Simplify the expression to find the second solution.

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Step 6.1

Subtract $2 π$ from $2 π + \frac{π}{4} + π$ .

$x = 2 π + \frac{π}{4} + π - 2 π$

Step 6.2

The resulting angle of $\frac{5 π}{4}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{4} + π$ .

$x = \frac{5 π}{4}$

Step 7

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

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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 $1$ 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

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

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Step 8.1

Add $2 π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + 2 π$

Step 8.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$2 π \cdot \frac{4}{4} - \frac{π}{4}$

Step 8.3

Combine fractions.

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Step 8.3.1

Combine $2 π$ and $\frac{4}{4}$ .

$\frac{2 π \cdot 4}{4} - \frac{π}{4}$

Step 8.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 4 - π}{4}$

Step 8.4

Simplify the numerator.

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Step 8.4.1

Multiply $4$ by $2$ .

$\frac{8 π - π}{4}$

Step 8.4.2

Subtract $π$ from $8 π$ .

$\frac{7 π}{4}$

Step 8.5

List the new angles.

$x = \frac{7 π}{4}$

Step 9

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

$x = \frac{5 π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$

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