Precalculus Examples

$\sec (θ) = \sqrt{2}$

Step 1

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

$θ = arcsec (\sqrt{2})$

Step 2

Simplify the right side.

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Step 2.1

The exact value of $arcsec (\sqrt{2})$ is $\frac{π}{4}$ .

$θ = \frac{π}{4}$

Step 3

The secant function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ = 2 π - \frac{π}{4}$

Step 4

Simplify $2 π - \frac{π}{4}$ .

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Step 4.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$θ = 2 π \cdot \frac{4}{4} - \frac{π}{4}$

Step 4.2

Combine fractions.

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Step 4.2.1

Combine $2 π$ and $\frac{4}{4}$ .

$θ = \frac{2 π \cdot 4}{4} - \frac{π}{4}$

Step 4.2.2

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 4 - π}{4}$

Step 4.3

Simplify the numerator.

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Step 4.3.1

Multiply $4$ by $2$ .

$θ = \frac{8 π - π}{4}$

Step 4.3.2

Subtract $π$ from $8 π$ .

$θ = \frac{7 π}{4}$

Step 5

Find the period of $\sec (θ)$ .

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Step 5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 5.2

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

$\frac{2 π}{| 1 |}$

Step 5.3

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

$\frac{2 π}{1}$

Step 5.4

Divide $2 π$ by $1$ .

$2 π$

Step 6

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

$θ = \frac{π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$