Trigonometry Examples

$\sec^{2} (θ) - 25 = 0$

Step 1

Add $25$ to both sides of the equation.

$\sec^{2} (θ) = 25$

Step 2

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

$\sec (θ) = \pm \sqrt{25}$

Step 3

Simplify $\pm \sqrt{25}$ .

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Step 3.1

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

$\sec (θ) = \pm \sqrt{5^{2}}$

Step 3.2

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

$\sec (θ) = \pm 5$

Step 4

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

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Step 4.1

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

$\sec (θ) = 5$

Step 4.2

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

$\sec (θ) = - 5$

Step 4.3

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

$\sec (θ) = 5, - 5$

Step 5

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

$\sec (θ) = 5$

$\sec (θ) = - 5$

Step 6

Solve for $θ$ in $\sec (θ) = 5$ .

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Step 6.1

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

$θ = arcsec (5)$

Step 6.2

Simplify the right side.

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Step 6.2.1

Evaluate $arcsec (5)$ .

$θ = 78.46304096$

Step 6.3

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.

$θ = 360 - 78.46304096$

Step 6.4

Subtract $78.46304096$ from $360$ .

$θ = 281.53695903$

Step 6.5

Find the period of $\sec (θ)$ .

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Step 6.5.1

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

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

Step 6.5.2

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

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

Step 6.5.3

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

$\frac{360}{1}$

Step 6.5.4

Divide $360$ by $1$ .

$360$

Step 6.6

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 $n$

Step 7

Solve for $θ$ in $\sec (θ) = - 5$ .

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Step 7.1

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

$θ = arcsec (- 5)$

Step 7.2

Simplify the right side.

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Step 7.2.1

Evaluate $arcsec (- 5)$ .

$θ = 101.53695903$

Step 7.3

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.

$θ = 360 - 101.53695903$

Step 7.4

Subtract $101.53695903$ from $360$ .

$θ = 258.46304096$

Step 7.5

Find the period of $\sec (θ)$ .

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Step 7.5.1

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

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

Step 7.5.2

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

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

Step 7.5.3

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

$\frac{360}{1}$

Step 7.5.4

Divide $360$ by $1$ .

$360$

Step 7.6

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 $n$

Step 8

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 $n$

Step 9

Consolidate the solutions.

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Step 9.1

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 $n$

Step 9.2

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 $n$