Solve for ? sin(x)=( square root of 3)/2

$\sin (x) = \frac{\sqrt{3}}{2}$

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

$x = \arcsin (\frac{\sqrt{3}}{2})$

The exact value of $\arcsin (\frac{\sqrt{3}}{2})$ is $\frac{π}{3}$ .

$x = \frac{π}{3}$

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.

$x = π - \frac{π}{3}$

Simplify $π - \frac{π}{3}$ .

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To write $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = \frac{π}{1} \cdot \frac{3}{3} - \frac{π}{3}$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{π \cdot 3}{1 \cdot 3} - \frac{π}{3}$

Multiply $3$ by $1$ .

$x = \frac{π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$x = \frac{π \cdot 3 - π}{3}$

Simplify the numerator.

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Move $3$ to the left of $π$ .

$x = \frac{3 \cdot π - π}{3}$

Subtract $π$ from $3 π$ .

$x = \frac{2 π}{3}$

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 $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n$ , for any integer $n$

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