Solve for θ 1-sin(theta)=1/2

$1 - \sin (θ) = \frac{1}{2}$

Step 1

Move all terms not containing $\sin (θ)$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$- \sin (θ) = \frac{1}{2} - 1$

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

$- \sin (θ) = \frac{1}{2} - 1 \cdot \frac{2}{2}$

Combine $- 1$ and $\frac{2}{2}$ .

$- \sin (θ) = \frac{1}{2} + \frac{- 1 \cdot 2}{2}$

Combine the numerators over the common denominator.

$- \sin (θ) = \frac{1 - 1 \cdot 2}{2}$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$- \sin (θ) = \frac{1 - 2}{2}$

Subtract $2$ from $1$ .

$- \sin (θ) = \frac{- 1}{2}$

Move the negative in front of the fraction.

$- \sin (θ) = - \frac{1}{2}$

Step 2

Divide each term in $- \sin (θ) = - \frac{1}{2}$ by $- 1$ and simplify.

Divide each term in $- \sin (θ) = - \frac{1}{2}$ by $- 1$ .

$\frac{- \sin (θ)}{- 1} = \frac{- \frac{1}{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\sin (θ)}{1} = \frac{- \frac{1}{2}}{- 1}$

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{- \frac{1}{2}}{- 1}$

Simplify the right side.

Dividing two negative values results in a positive value.

$\sin (θ) = \frac{\frac{1}{2}}{1}$

Divide $\frac{1}{2}$ by $1$ .

$\sin (θ) = \frac{1}{2}$

Step 3

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

$θ = \arcsin (\frac{1}{2})$

Step 4

Simplify the right side.

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$θ = \frac{π}{6}$

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.

$θ = π - \frac{π}{6}$

Step 6

Simplify $π - \frac{π}{6}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$θ = π \cdot \frac{6}{6} - \frac{π}{6}$

Combine fractions.

Combine $π$ and $\frac{6}{6}$ .

$θ = \frac{π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 6 - π}{6}$

Simplify the numerator.

Move $6$ to the left of $π$ .

$θ = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{6}$

Step 7

Find the period of $\sin (θ)$ .

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.

$\frac{2 π}{| 1 |}$

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 $1$ .

$2 π$

Step 8

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

$θ = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Solve for θ 1-sin(theta)=1/2

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