Solve for θ sec(theta)=(2 square root of 3)/3

$\sec (θ) = \frac{2 \sqrt{3}}{3}$

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

$θ = arcsec (\frac{2 \sqrt{3}}{3})$

Simplify the right side.

The exact value of $arcsec (\frac{2 \sqrt{3}}{3})$ is $\frac{π}{6}$ .

$θ = \frac{π}{6}$

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.

$θ = 2 π - \frac{π}{6}$

Simplify $2 π - \frac{π}{6}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$θ = 2 π \cdot \frac{6}{6} - \frac{π}{6}$

Combine fractions.

Combine $2 π$ and $\frac{6}{6}$ .

$θ = \frac{2 π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 6 - π}{6}$

Simplify the numerator.

Multiply $6$ by $2$ .

$θ = \frac{12 π - π}{6}$

Subtract $π$ from $12 π$ .

$θ = \frac{11 π}{6}$

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 $1$ .

$2 π$

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

$θ = \frac{π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Solve for θ sec(theta)=(2 square root of 3)/3

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