Trigonometry Examples

$\sec (θ) = 1$

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.

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Step 2.1

The exact value of $arcsec (1)$ is $0$ .

$θ = 0$

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.

$θ = 2 π - 0$

Step 4

Subtract $0$ from $2 π$ .

$θ = 2 π$

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.

$θ = 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 7

Consolidate the answers.

$θ = 2 π n$ , for any integer $n$