Solve for θ in Degrees sec(theta)^2-25=0

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Trigonometry

$\sec^{2} (θ) - 25 = 0$

Add $25$ to both sides of the equation.

$\sec^{2} (θ) = 25$

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

$\sec (θ) = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

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Rewrite $25$ as $5^{2}$ .

$\sec (θ) = \pm \sqrt{5^{2}}$

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

$\sec (θ) = \pm 5$

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.

$\sec (θ) = 5$

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

$\sec (θ) = - 5$

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

$\sec (θ) = 5, - 5$

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

$\sec (θ) = 5$

$\sec (θ) = - 5$

Solve for $θ$ in $\sec (θ) = 5$ .

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

$θ = arcsec (5)$

Simplify the right side.

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Evaluate $arcsec (5)$ .

$θ = 78.46304096$

The secant function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the fourth quadrant.

$θ = 360 - 78.46304096$

Subtract $78.46304096$ from $360$ .

$θ = 281.53695903$

Find the period of $\sec (θ)$ .

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

$θ = 78.46304096 + 360 n, 281.53695903 + 360 n$ , for any integer $n$

Solve for $θ$ in $\sec (θ) = - 5$ .

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

$θ = arcsec (- 5)$

Simplify the right side.

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Evaluate $arcsec (- 5)$ .

$θ = 101.53695903$

The secant function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the third quadrant.

$θ = 360 - 101.53695903$

Subtract $101.53695903$ from $360$ .

$θ = 258.46304096$

Find the period of $\sec (θ)$ .

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

$θ = 101.53695903 + 360 n, 258.46304096 + 360 n$ , for any integer $n$

List all of the solutions.

$θ = 78.46304096 + 360 n, 281.53695903 + 360 n, 101.53695903 + 360 n, 258.46304096 + 360 n$ , for any integer $n$

Consolidate the solutions.

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Consolidate $78.46304096 + 360 n$ and $258.46304096 + 360 n$ to $78.46304096 + 180 n$ .

$θ = 78.46304096 + 180 n, 281.53695903 + 360 n, 101.53695903 + 360 n$ , for any integer $n$

Consolidate $281.53695903 + 360 n$ and $101.53695903 + 360 n$ to $101.53695903 + 180 n$ .

$θ = 78.46304096 + 180 n, 101.53695903 + 180 n$ , for any integer $n$

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