Solve for θ sec(theta)=-1

$\sec (θ) = - 1$

Step 1

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

$θ = arcsec (- 1)$

Step 2

Simplify the right side.

The exact value of $arcsec (- 1)$ is $π$ .

$θ = π$

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.

$θ = 2 π - π$

Step 4

Subtract $π$ from $2 π$ .

$θ = π$

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.

$θ = π + 2 π n$ , for any integer $n$

Solve for θ sec(theta)=-1

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