Trigonometry Examples

$2 \sin (3 x + 15) - 5 = - 6$

Step 1

Move all terms not containing $\sin (3 x + 15)$ to the right side of the equation.

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Step 1.1

Add $5$ to both sides of the equation.

$2 \sin (3 x + 15) = - 6 + 5$

Step 1.2

Add $- 6$ and $5$ .

$2 \sin (3 x + 15) = - 1$

Step 2

Divide each term in $2 \sin (3 x + 15) = - 1$ by $2$ and simplify.

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Step 2.1

Divide each term in $2 \sin (3 x + 15) = - 1$ by $2$ .

$\frac{2 \sin (3 x + 15)}{2} = \frac{- 1}{2}$

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.

$\frac{2 \sin (3 x + 15)}{2} = \frac{- 1}{2}$

Step 2.2.1.2

Divide $\sin (3 x + 15)$ by $1$ .

$\sin (3 x + 15) = \frac{- 1}{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Move the negative in front of the fraction.

$\sin (3 x + 15) = - \frac{1}{2}$

Step 3

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

$3 x + 15 = \arcsin (- \frac{1}{2})$

Step 4

Simplify the right side.

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Step 4.1

The exact value of $\arcsin (- \frac{1}{2})$ is $- \frac{π}{6}$ .

$3 x + 15 = - \frac{π}{6}$

Step 5

Subtract $15$ from both sides of the equation.

$3 x = - \frac{π}{6} - 15$

Step 6

Divide each term in $3 x = - \frac{π}{6} - 15$ by $3$ and simplify.

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Step 6.1

Divide each term in $3 x = - \frac{π}{6} - 15$ by $3$ .

$\frac{3 x}{3} = \frac{- \frac{π}{6}}{3} + \frac{- 15}{3}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $3$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- \frac{π}{6}}{3} + \frac{- 15}{3}$

Step 6.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \frac{π}{6}}{3} + \frac{- 15}{3}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Simplify each term.

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Step 6.3.1.1

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{π}{6} \cdot \frac{1}{3} + \frac{- 15}{3}$

Step 6.3.1.2

Multiply $- \frac{π}{6} \cdot \frac{1}{3}$ .

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Step 6.3.1.2.1

Multiply $\frac{1}{3}$ by $\frac{π}{6}$ .

$x = - \frac{π}{3 \cdot 6} + \frac{- 15}{3}$

Step 6.3.1.2.2

Multiply $3$ by $6$ .

$x = - \frac{π}{18} + \frac{- 15}{3}$

Step 6.3.1.3

Divide $- 15$ by $3$ .

$x = - \frac{π}{18} - 5$

Step 7

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.

$3 x + 15 = 2 π + \frac{π}{6} + π$

Step 8

Simplify the expression to find the second solution.

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Step 8.1

Subtract $2 π$ from $2 π + \frac{π}{6} + π$ .

$3 x + 15 = 2 π + \frac{π}{6} + π - 2 π$

Step 8.2

The resulting angle of $\frac{7 π}{6}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{6} + π$ .

$3 x + 15 = \frac{7 π}{6}$

Step 8.3

Solve for $x$ .

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Step 8.3.1

Subtract $15$ from both sides of the equation.

$3 x = \frac{7 π}{6} - 15$

Step 8.3.2

Divide each term in $3 x = \frac{7 π}{6} - 15$ by $3$ and simplify.

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Step 8.3.2.1

Divide each term in $3 x = \frac{7 π}{6} - 15$ by $3$ .

$\frac{3 x}{3} = \frac{\frac{7 π}{6}}{3} + \frac{- 15}{3}$

Step 8.3.2.2

Simplify the left side.

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Step 8.3.2.2.1

Cancel the common factor of $3$ .

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Step 8.3.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{\frac{7 π}{6}}{3} + \frac{- 15}{3}$

Step 8.3.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{7 π}{6}}{3} + \frac{- 15}{3}$

Step 8.3.2.3

Simplify the right side.

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Step 8.3.2.3.1

Simplify each term.

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Step 8.3.2.3.1.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{7 π}{6} \cdot \frac{1}{3} + \frac{- 15}{3}$

Step 8.3.2.3.1.2

Multiply $\frac{7 π}{6} \cdot \frac{1}{3}$ .

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Step 8.3.2.3.1.2.1

Multiply $\frac{7 π}{6}$ by $\frac{1}{3}$ .

$x = \frac{7 π}{6 \cdot 3} + \frac{- 15}{3}$

Step 8.3.2.3.1.2.2

Multiply $6$ by $3$ .

$x = \frac{7 π}{18} + \frac{- 15}{3}$

Step 8.3.2.3.1.3

Divide $- 15$ by $3$ .

$x = \frac{7 π}{18} - 5$

Step 9

Find the period of $\sin (3 x + 15)$ .

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Step 9.1

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

$\frac{2 π}{| b |}$

Step 9.2

Replace $b$ with $3$ in the formula for period.

$\frac{2 π}{| 3 |}$

Step 9.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$\frac{2 π}{3}$

Step 10

Add $\frac{2 π}{3}$ to every negative angle to get positive angles.

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Step 10.1

Add $\frac{2 π}{3}$ to $- \frac{π}{18} - 5$ to find the positive angle.

$- \frac{π}{18} - 5 + \frac{2 π}{3}$

Step 10.2

To write $\frac{2 π}{3}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$\frac{2 π}{3} \cdot \frac{6}{6} - \frac{π}{18} - 5$

Step 10.3

Write each expression with a common denominator of $18$ , by multiplying each by an appropriate factor of $1$ .

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Step 10.3.1

Multiply $\frac{2 π}{3}$ by $\frac{6}{6}$ .

$\frac{2 π \cdot 6}{3 \cdot 6} - \frac{π}{18} - 5$

Step 10.3.2

Multiply $3$ by $6$ .

$\frac{2 π \cdot 6}{18} - \frac{π}{18} - 5$

Step 10.4

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{18} - 5$

Step 10.5

Simplify the numerator.

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Step 10.5.1

Multiply $6$ by $2$ .

$\frac{12 π - π}{18} - 5$

Step 10.5.2

Subtract $π$ from $12 π$ .

$\frac{11 π}{18} - 5$

Step 10.6

List the new angles.

$x = \frac{11 π}{18} - 5$

Step 11

The period of the $\sin (3 x + 15)$ function is $\frac{2 π}{3}$ so values will repeat every $\frac{2 π}{3}$ radians in both directions.

$x = \frac{7 π}{18} - 5 + \frac{2 π n}{3}, \frac{11 π}{18} - 5 + \frac{2 π n}{3}$ , for any integer $n$