Solve for θ sin(theta/2)=-1

$\sin (\frac{θ}{2}) = - 1$

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

$\frac{θ}{2} = \arcsin (- 1)$

Simplify the right side.

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

$\frac{θ}{2} = - \frac{π}{2}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$θ = - π$

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.

$\frac{θ}{2} = 2 π + \frac{π}{2} + π$

Simplify the expression to find the second solution.

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

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

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

$\frac{θ}{2} = \frac{3 π}{2}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$θ = 3 π$

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 $- π$ to find the positive angle.

$- π + 4 π$

Subtract $π$ from $4 π$ .

$3 π$

List the new angles.

$θ = 3 π$

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

$θ = 3 π + 4 π n, 3 π + 4 π n$ , for any integer $n$

Consolidate the answers.

$θ = 3 π + 4 π n$ , for any integer $n$

Solve for θ sin(theta/2)=-1

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