Solve for ? sin(x) equals one First, divide two

$\sin (x) = \frac{1}{2}$

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

$x = \arcsin (\frac{1}{2})$

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

$x = \frac{π}{6}$

The 1st and 2nd quadrants of the sine function are positive. To calculate the second answer, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - \frac{π}{6}$

Clarify $π - \frac{π}{6}$ .

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To write $\frac{π}{1}$ as a fraction with a common denominator, First, multiply by $\frac{6}{6}$ .

$x = \frac{π}{1} \cdot \frac{6}{6} - \frac{π}{6}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Add.

$x = \frac{π \cdot 6}{1 \cdot 6} - \frac{π}{6}$

Multiply $6$ by $1$ .

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

Add numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Clarify the numerator.

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Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Calculate the period.

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Evaluate the period of function using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Find solution to 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 $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in two directions at once.

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

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