Trigonometry Examples

$16 \sec^{2} (θ) - 25 = 0$

Step 1

Add $25$ to both sides of the equation.

$16 \sec^{2} (θ) = 25$

Step 2

Divide each term in $16 \sec^{2} (θ) = 25$ by $16$ and simplify.

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Step 2.1

Divide each term in $16 \sec^{2} (θ) = 25$ by $16$ .

$\frac{16 \sec^{2} (θ)}{16} = \frac{25}{16}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $16$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{16 \sec^{2} (θ)}{16} = \frac{25}{16}$

Step 2.2.1.2

Divide $\sec^{2} (θ)$ by $1$ .

$\sec^{2} (θ) = \frac{25}{16}$

Step 3

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

$\sec (θ) = \pm \sqrt{\frac{25}{16}}$

Step 4

Simplify $\pm \sqrt{\frac{25}{16}}$ .

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Step 4.1

Rewrite $\sqrt{\frac{25}{16}}$ as $\frac{\sqrt{25}}{\sqrt{16}}$ .

$\sec (θ) = \pm \frac{\sqrt{25}}{\sqrt{16}}$

Step 4.2

Simplify the numerator.

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Step 4.2.1

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

$\sec (θ) = \pm \frac{\sqrt{5^{2}}}{\sqrt{16}}$

Step 4.2.2

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

$\sec (θ) = \pm \frac{5}{\sqrt{16}}$

Step 4.3

Simplify the denominator.

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Step 4.3.1

Rewrite $16$ as $4^{2}$ .

$\sec (θ) = \pm \frac{5}{\sqrt{4^{2}}}$

Step 4.3.2

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

$\sec (θ) = \pm \frac{5}{4}$

Step 5

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

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Step 5.1

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

$\sec (θ) = \frac{5}{4}$

Step 5.2

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

$\sec (θ) = - \frac{5}{4}$

Step 5.3

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

$\sec (θ) = \frac{5}{4}, - \frac{5}{4}$

Step 6

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

$\sec (θ) = \frac{5}{4}$

$\sec (θ) = - \frac{5}{4}$

Step 7

Solve for $θ$ in $\sec (θ) = \frac{5}{4}$ .

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Step 7.1

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

$θ = arcsec (\frac{5}{4})$

Step 7.2

Simplify the right side.

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Step 7.2.1

Evaluate $arcsec (\frac{5}{4})$ .

$θ = 36.86989764$

Step 7.3

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.

$θ = 360 - 36.86989764$

Step 7.4

Subtract $36.86989764$ from $360$ .

$θ = 323.13010235$

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.

$θ = 36.86989764 + 360 n, 323.13010235 + 360 n$ , for any integer $n$

Step 8

Solve for $θ$ in $\sec (θ) = - \frac{5}{4}$ .

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Step 8.1

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

$θ = arcsec (- \frac{5}{4})$

Step 8.2

Simplify the right side.

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Step 8.2.1

Evaluate $arcsec (- \frac{5}{4})$ .

$θ = 143.13010235$

Step 8.3

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.

$θ = 360 - 143.13010235$

Step 8.4

Subtract $143.13010235$ from $360$ .

$θ = 216.86989764$

Step 8.5

Find the period of $\sec (θ)$ .

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Step 8.5.1

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

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

Step 8.5.2

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

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

Step 8.5.3

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

$\frac{360}{1}$

Step 8.5.4

Divide $360$ by $1$ .

$360$

Step 8.6

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

$θ = 143.13010235 + 360 n, 216.86989764 + 360 n$ , for any integer $n$

Step 9

List all of the solutions.

$θ = 36.86989764 + 360 n, 323.13010235 + 360 n, 143.13010235 + 360 n, 216.86989764 + 360 n$ , for any integer $n$

Step 10

Consolidate the solutions.

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Step 10.1

Consolidate $36.86989764 + 360 n$ and $216.86989764 + 360 n$ to $36.86989764 + 180 n$ .

$θ = 36.86989764 + 180 n, 323.13010235 + 360 n, 143.13010235 + 360 n$ , for any integer $n$

Step 10.2

Consolidate $323.13010235 + 360 n$ and $143.13010235 + 360 n$ to $143.13010235 + 180 n$ .

$θ = 36.86989764 + 180 n, 143.13010235 + 180 n$ , for any integer $n$