Solve for θ sin(theta/2)-1=0

$\sin (\frac{θ}{2}) - 1 = 0$

Step 1

Add $1$ to both sides of the equation.

$\sin (\frac{θ}{2}) = 1$

Step 2

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

$\frac{θ}{2} = \arcsin (1)$

Step 3

Simplify the right side.

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

$\frac{θ}{2} = \frac{π}{2}$

Step 4

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

$θ = π$

Step 5

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$\frac{θ}{2} = π - \frac{π}{2}$

Step 6

Solve for $θ$ .

Multiply both sides of the equation by $2$ .

$2 \frac{θ}{2} = 2 (π - \frac{π}{2})$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{θ}{2} = 2 (π - \frac{π}{2})$

Rewrite the expression.

$θ = 2 (π - \frac{π}{2})$

Simplify the right side.

Simplify $2 (π - \frac{π}{2})$ .

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

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

Simplify terms.

Combine $π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$θ = 2 \frac{π \cdot 2 - π}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$θ = 2 \frac{π \cdot 2 - π}{2}$

Rewrite the expression.

$θ = π \cdot 2 - π$

Move $2$ to the left of $π$ .

$θ = 2 π - π$

Subtract $π$ from $2 π$ .

$θ = π$

Step 7

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 $2$ .

$4 π$

Step 8

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

$θ = π + 4 π n$ , for any integer $n$

Solve for θ sin(theta/2)-1=0

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