Solve for θ 4sin(theta)^2=1

$4 \sin^{2} (θ) = 1$

Step 1

Divide each term in $4 \sin^{2} (θ) = 1$ by $4$ and simplify.

Divide each term in $4 \sin^{2} (θ) = 1$ by $4$ .

$\frac{4 \sin^{2} (θ)}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 \sin^{2} (θ)}{4} = \frac{1}{4}$

Divide $\sin^{2} (θ)$ by $1$ .

$\sin^{2} (θ) = \frac{1}{4}$

Step 2

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$\sin (θ) = \pm \sqrt{\frac{1}{4}}$

Step 3

Simplify $\pm \sqrt{\frac{1}{4}}$ .

Rewrite $\sqrt{\frac{1}{4}}$ as $\frac{\sqrt{1}}{\sqrt{4}}$ .

$\sin (θ) = \pm \frac{\sqrt{1}}{\sqrt{4}}$

Any root of $1$ is $1$ .

$\sin (θ) = \pm \frac{1}{\sqrt{4}}$

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$\sin (θ) = \pm \frac{1}{\sqrt{2^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$\sin (θ) = \pm \frac{1}{2}$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$\sin (θ) = \frac{1}{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$\sin (θ) = - \frac{1}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$\sin (θ) = \frac{1}{2}, - \frac{1}{2}$

Step 5

Set up each of the solutions to solve for $θ$ .

$\sin (θ) = \frac{1}{2}$

$\sin (θ) = - \frac{1}{2}$

Step 6

Solve for $θ$ in $\sin (θ) = \frac{1}{2}$ .

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

$θ = \arcsin (\frac{1}{2})$

Simplify the right side.

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

$θ = \frac{π}{6}$

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.

$θ = π - \frac{π}{6}$

Simplify $π - \frac{π}{6}$ .

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

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

Combine fractions.

Combine $π$ and $\frac{6}{6}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 6 - π}{6}$

Simplify the numerator.

Move $6$ to the left of $π$ .

$θ = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{6}$

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 $1$ .

$2 π$

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

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

Step 7

Solve for $θ$ in $\sin (θ) = - \frac{1}{2}$ .

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

$θ = \arcsin (- \frac{1}{2})$

Simplify the right side.

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

$θ = - \frac{π}{6}$

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$θ = 2 π + \frac{π}{6} + π$

Simplify the expression to find the second solution.

Subtract $2 π$ from $2 π + \frac{π}{6} + π$ .

$θ = 2 π + \frac{π}{6} + π - 2 π$

The resulting angle of $\frac{7 π}{6}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{6} + π$ .

$θ = \frac{7 π}{6}$

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 $1$ .

$2 π$

Add $2 π$ to every negative angle to get positive angles.

Add $2 π$ to $- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + 2 π$

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$2 π \cdot \frac{6}{6} - \frac{π}{6}$

Combine fractions.

Combine $2 π$ and $\frac{6}{6}$ .

$\frac{2 π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{6}$

Simplify the numerator.

Multiply $6$ by $2$ .

$\frac{12 π - π}{6}$

Subtract $π$ from $12 π$ .

$\frac{11 π}{6}$

List the new angles.

$θ = \frac{11 π}{6}$

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

$θ = \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Step 8

List all of the solutions.

$θ = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Step 9

Consolidate the solutions.

Consolidate $\frac{π}{6} + 2 π n$ and $\frac{7 π}{6} + 2 π n$ to $\frac{π}{6} + π n$ .

$θ = \frac{π}{6} + π n, \frac{5 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Consolidate $\frac{5 π}{6} + 2 π n$ and $\frac{11 π}{6} + 2 π n$ to $\frac{5 π}{6} + π n$ .

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

Solve for θ 4sin(theta)^2=1

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