Evaluate sin(3theta)=-1

$\sin (3 θ) = - 1$

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

$3 θ = \arcsin (- 1)$

Simplify the right side.

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

$3 θ = - \frac{π}{2}$

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

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

$\frac{3 θ}{3} = \frac{- \frac{π}{2}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 θ}{3} = \frac{- \frac{π}{2}}{3}$

Divide $θ$ by $1$ .

$θ = \frac{- \frac{π}{2}}{3}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = - \frac{π}{2} \cdot \frac{1}{3}$

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

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

$θ = - \frac{π}{3 \cdot 2}$

Multiply $3$ by $2$ .

$θ = - \frac{π}{6}$

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.

$3 θ = 2 π + \frac{π}{2} + π$

Simplify the expression to find the second solution.

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

$3 θ = 2 π + \frac{π}{2} + π - 2 π$

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

$3 θ = \frac{3 π}{2}$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $θ$ by $1$ .

$θ = \frac{\frac{3 π}{2}}{3}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{3 π}{2} \cdot \frac{1}{3}$

Cancel the common factor of $3$ .

Factor $3$ out of $3 π$ .

$θ = \frac{3 (π)}{2} \cdot \frac{1}{3}$

Cancel the common factor.

$θ = \frac{3 π}{2} \cdot \frac{1}{3}$

Rewrite the expression.

$θ = \frac{π}{2}$

Find the period of $\sin (3 θ)$ .

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 3 |}$

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

$\frac{2 π}{3}$

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

Add $\frac{2 π}{3}$ to $- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + \frac{2 π}{3}$

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

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

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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

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

Multiply $3$ by $2$ .

$\frac{2 π \cdot 2}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{6}$

Simplify the numerator.

Multiply $2$ by $2$ .

$\frac{4 π - π}{6}$

Subtract $π$ from $4 π$ .

$\frac{3 π}{6}$

Cancel the common factor of $3$ and $6$ .

Factor $3$ out of $3 π$ .

$\frac{3 (π)}{6}$

Cancel the common factors.

Factor $3$ out of $6$ .

$\frac{3 π}{3 \cdot 2}$

Cancel the common factor.

$\frac{3 π}{3 \cdot 2}$

Rewrite the expression.

$\frac{π}{2}$

List the new angles.

$θ = \frac{π}{2}$

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

$θ = \frac{π}{2} + \frac{2 π n}{3}, \frac{π}{2} + \frac{2 π n}{3}$ , for any integer $n$

Consolidate the answers.

$θ = \frac{π}{2} + \frac{2 π n}{3}$ , for any integer $n$

Evaluate sin(3theta)=-1

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