Solve for θ 2sin(theta) = square root of 2

$2 \sin (θ) = \sqrt{2}$

Step 1

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

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

$\frac{2 \sin (θ)}{2} = \frac{\sqrt{2}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \sin (θ)}{2} = \frac{\sqrt{2}}{2}$

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{\sqrt{2}}{2}$

Step 2

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

$θ = \arcsin (\frac{\sqrt{2}}{2})$

Step 3

Simplify the right side.

The exact value of $\arcsin (\frac{\sqrt{2}}{2})$ is $\frac{π}{4}$ .

$θ = \frac{π}{4}$

Step 4

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.

$θ = π - \frac{π}{4}$

Step 5

Simplify $π - \frac{π}{4}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$θ = π \cdot \frac{4}{4} - \frac{π}{4}$

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

$θ = \frac{π \cdot 4}{4} - \frac{π}{4}$

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 4 - π}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$θ = \frac{4 \cdot π - π}{4}$

Subtract $π$ from $4 π$ .

$θ = \frac{3 π}{4}$

Step 6

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 7

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

$θ = \frac{π}{4} + 2 π n, \frac{3 π}{4} + 2 π n$ , for any integer $n$

Solve for θ 2sin(theta) = square root of 2

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