Solve for x in Degrees 2sin(x)=1

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Trigonometry

$2 \sin (x) = 1$

Divide each term in $2 \sin (x) = 1$ by $2$ and simplify.

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Divide each term in $2 \sin (x) = 1$ by $2$ .

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

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $\sin (x)$ by $1$ .

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

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

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

Simplify the right side.

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The exact value of $\arcsin (\frac{1}{2})$ is $30$ .

$x = 30$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the second quadrant.

$x = 180 - 30$

Subtract $30$ from $180$ .

$x = 150$

Find the period of $\sin (x)$ .

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

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

$\frac{360}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

The period of the $\sin (x)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$x = 30 + 360 n, 150 + 360 n$ , for any integer $n$

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