Solve for ? sin(x)=35/37

$\sin (x) = \frac{35}{37}$

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

$x = \arcsin (\frac{35}{37})$

Evaluate $\arcsin (\frac{35}{37})$ .

$x = 1.24049897$

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.

$x = (3.14159265) - 1.24049897$

Simplify the expression to find the second solution.

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Remove the parentheses around the expression $3.14159265$ .

$x = 3.14159265 - 1.24049897$

Subtract $1.24049897$ from $3.14159265$ .

$x = 1.90109368$

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.

$x = 1.24049897 + 2 π n, 1.90109368 + 2 π n$ , for any integer $n$

Solve for ? sin(x)=35/37

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