Solve for x sin(x)=-1

$\sin (x) = - 1$

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

$x = \arcsin (- 1)$

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

$x = - \frac{π}{2}$

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.

$x = 2 π + \frac{π}{2} + π$

Simplify the expression to find the second solution.

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Simplify $2 π + \frac{π}{2} + π$ .

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To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{2 π}{1} \cdot \frac{2}{2} + \frac{π}{2} + π$

Write each expression with a common denominator of $2$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{2 π \cdot 2}{1 \cdot 2} + \frac{π}{2} + π$

Multiply $2$ by $1$ .

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

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 + π}{2} + π$

Simplify the numerator.

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Multiply $2$ by $2$ .

$x = \frac{4 π + π}{2} + π$

Add $4 π$ and $π$ .

$x = \frac{5 π}{2} + π$

To write $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{5 π}{2} + \frac{π}{1} \cdot \frac{2}{2}$

Write each expression with a common denominator of $2$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{5 π}{2} + \frac{π \cdot 2}{1 \cdot 2}$

Multiply $2$ by $1$ .

$x = \frac{5 π}{2} + \frac{π \cdot 2}{2}$

Combine the numerators over the common denominator.

$x = \frac{5 π + π \cdot 2}{2}$

Simplify the numerator.

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Move $2$ to the left of $π$ .

$x = \frac{5 π + 2 \cdot π}{2}$

Add $5 π$ and $2 π$ .

$x = \frac{7 π}{2}$

Subtract $2 π$ from $\frac{7 π}{2}$ .

$x = \frac{7 π}{2} - 2 π$

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $\frac{7 π}{2}$ .

$x = \frac{3 π}{2}$

Find the period.

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

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Solve the equation.

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The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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

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

$- \frac{π}{2} + 2 π$

To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Write each expression with a common denominator of $2$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

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

Multiply $2$ by $1$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

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Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

List the new angles.

$x = \frac{3 π}{2}$

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

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

Consolidate the answers.

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

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