Trigonometry Examples

$\sec (x) - 2 = 0$

Step 1

Add $2$ to both sides of the equation.

$\sec (x) = 2$

Step 2

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

$x = arcsec (2)$

Step 3

Simplify the right side.

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Step 3.1

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

$x = \frac{π}{3}$

Step 4

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.

$x = 2 π - \frac{π}{3}$

Step 5

Simplify $2 π - \frac{π}{3}$ .

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Step 5.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = 2 π \cdot \frac{3}{3} - \frac{π}{3}$

Step 5.2

Combine fractions.

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Step 5.2.1

Combine $2 π$ and $\frac{3}{3}$ .

$x = \frac{2 π \cdot 3}{3} - \frac{π}{3}$

Step 5.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 5.3

Simplify the numerator.

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Step 5.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 5.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 6

Find the period of $\sec (x)$ .

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Step 6.1

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

$\frac{2 π}{| b |}$

Step 6.2

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

$\frac{2 π}{| 1 |}$

Step 6.3

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

$\frac{2 π}{1}$

Step 6.4

Divide $2 π$ by $1$ .

$2 π$

Step 7

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

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