Solve for θ in Degrees 4sin(theta)^2=1

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Trigonometry

$4 \sin^{2} (θ) = 1$

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

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

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

Simplify the left side.

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Cancel the common factor of $4$ .

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Cancel the common factor.

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

Divide $\sin^{2} (θ)$ by $1$ .

$\sin^{2} (θ) = \frac{1}{4}$

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

$\sin (θ) = \pm \sqrt{\frac{1}{4}}$

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

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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.

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

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

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

$\sin (θ) = \pm \frac{1}{2}$

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

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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.

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

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

$\sin (θ) = \frac{1}{2}$

$\sin (θ) = - \frac{1}{2}$

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

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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.

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

$θ = 30$

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.

$θ = 180 - 30$

Subtract $30$ from $180$ .

$θ = 150$

Find the period of $\sin (θ)$ .

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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 $\sin (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

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

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

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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.

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

$θ = - 30$

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $360$ , to find a reference angle. Next, add this reference angle to $180$ to find the solution in the third quadrant.

$θ = 360 + 30 + 180$

Simplify the expression to find the second solution.

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Subtract $360 °$ from $360 + 30 + 180 °$ .

$θ = 360 + 30 + 180 ° - 360 °$

The resulting angle of $210 °$ is positive, less than $360 °$ , and coterminal with $360 + 30 + 180$ .

$θ = 210 °$

Find the period of $\sin (θ)$ .

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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 $1$ .

$360$

Add $360$ to every negative angle to get positive angles.

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Add $360$ to $- 30$ to find the positive angle.

$- 30 + 360$

Subtract $30$ from $360$ .

$330$

List the new angles.

$θ = 330$

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

$θ = 210 + 360 n, 330 + 360 n$ , for any integer $n$

List all of the solutions.

$θ = 30 + 360 n, 150 + 360 n, 210 + 360 n, 330 + 360 n$ , for any integer $n$

Consolidate the solutions.

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Consolidate $30 + 360 n$ and $210 + 360 n$ to $30 + 180 n$ .

$θ = 30 + 180 n, 150 + 360 n, 330 + 360 n$ , for any integer $n$

Consolidate $150 + 360 n$ and $330 + 360 n$ to $150 + 180 n$ .

$θ = 30 + 180 n, 150 + 180 n$ , for any integer $n$

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