Solve for θ sec(theta)=- square root of 2

$\sec (θ) = - \sqrt{2}$

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.

The exact value of $arcsec (- \sqrt{2})$ is $\frac{3 π}{4}$ .

$θ = \frac{3 π}{4}$

Step 3

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

$θ = 2 π - \frac{3 π}{4}$

Step 4

Simplify $2 π - \frac{3 π}{4}$ .

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

$θ = 2 π \cdot \frac{4}{4} - \frac{3 π}{4}$

Combine fractions.

Combine $2 π$ and $\frac{4}{4}$ .

$θ = \frac{2 π \cdot 4}{4} - \frac{3 π}{4}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 4 - 3 π}{4}$

Simplify the numerator.

Multiply $4$ by $2$ .

$θ = \frac{8 π - 3 π}{4}$

Subtract $3 π$ from $8 π$ .

$θ = \frac{5 π}{4}$

Step 5

Find the period of $\sec (θ)$ .

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

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.

$θ = \frac{3 π}{4} + 2 π n, \frac{5 π}{4} + 2 π n$ , for any integer $n$

Solve for θ sec(theta)=- square root of 2

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