Solve for ? sin(theta)-1=0

$\sin (θ) - 1 = 0$

Add $1$ to both sides of the equation.

$\sin (θ) = 1$

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

$θ = \arcsin (1)$

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

$θ = \frac{π}{2}$

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.

$θ = π - \frac{π}{2}$

Simplify $π - \frac{π}{2}$ .

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

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

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

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Combine.

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

Multiply $2$ by $1$ .

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

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 2 - π}{2}$

Simplify the numerator.

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Move $2$ to the left of $π$ .

$θ = \frac{2 \cdot π - π}{2}$

Subtract $π$ from $2 π$ .

$θ = \frac{π}{2}$

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 $1$ .

$2 π$

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

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

Solve for ? sin(theta)-1=0

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