Solve for θ sin(theta/2)=-( square root of 3)/2

$\sin (\frac{θ}{2}) = - \frac{\sqrt{3}}{2}$

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

$\frac{θ}{2} = \arcsin (- \frac{\sqrt{3}}{2})$

Simplify the right side.

The exact value of $\arcsin (- \frac{\sqrt{3}}{2})$ is $- \frac{π}{3}$ .

$\frac{θ}{2} = - \frac{π}{3}$

Multiply both sides of the equation by $2$ .

$2 \frac{θ}{2} = 2 (- \frac{π}{3})$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{θ}{2} = 2 (- \frac{π}{3})$

Rewrite the expression.

$θ = 2 (- \frac{π}{3})$

Simplify the right side.

Simplify $2 (- \frac{π}{3})$ .

Multiply $2 (- \frac{π}{3})$ .

Multiply $- 1$ by $2$ .

$θ = - 2 \frac{π}{3}$

Combine $- 2$ and $\frac{π}{3}$ .

$θ = \frac{- 2 π}{3}$

Move the negative in front of the fraction.

$θ = - \frac{2 π}{3}$

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.

$\frac{θ}{2} = 2 π + \frac{π}{3} + π$

Simplify the expression to find the second solution.

Subtract $2 π$ from $2 π + \frac{π}{3} + π$ .

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

The resulting angle of $\frac{4 π}{3}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{3} + π$ .

$\frac{θ}{2} = \frac{4 π}{3}$

Solve for $θ$ .

Multiply both sides of the equation by $2$ .

$2 \frac{θ}{2} = 2 \frac{4 π}{3}$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{θ}{2} = 2 \frac{4 π}{3}$

Rewrite the expression.

$θ = 2 \frac{4 π}{3}$

Simplify the right side.

Multiply $2 \frac{4 π}{3}$ .

Combine $2$ and $\frac{4 π}{3}$ .

$θ = \frac{2 (4 π)}{3}$

Multiply $4$ by $2$ .

$θ = \frac{8 π}{3}$

Find the period of $\sin (\frac{θ}{2})$ .

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

$\frac{2 π}{| b |}$

Replace $b$ with $\frac{1}{2}$ in the formula for period.

$\frac{2 π}{| \frac{1}{2} |}$

$\frac{1}{2}$ is approximately $0.5$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{2}}$

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 2$

Multiply $2$ by $2$ .

$4 π$

Add $4 π$ to every negative angle to get positive angles.

Add $4 π$ to $- \frac{2 π}{3}$ to find the positive angle.

$- \frac{2 π}{3} + 4 π$

To write $4 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$4 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Combine fractions.

Combine $4 π$ and $\frac{3}{3}$ .

$\frac{4 π \cdot 3}{3} - \frac{2 π}{3}$

Combine the numerators over the common denominator.

$\frac{4 π \cdot 3 - 2 π}{3}$

Simplify the numerator.

Multiply $3$ by $4$ .

$\frac{12 π - 2 π}{3}$

Subtract $2 π$ from $12 π$ .

$\frac{10 π}{3}$

List the new angles.

$θ = \frac{10 π}{3}$

The period of the $\sin (\frac{θ}{2})$ function is $4 π$ so values will repeat every $4 π$ radians in both directions.

$θ = \frac{8 π}{3} + 4 π n, \frac{10 π}{3} + 4 π n$ , for any integer $n$

Solve for θ sin(theta/2)=-( square root of 3)/2

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