Trigonometry Examples

$\sec (3 x) = - 1$

Step 1

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

$3 x = arcsec (- 1)$

Step 2

Simplify the right side.

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Step 2.1

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

$3 x = π$

Step 3

Divide each term in $3 x = π$ by $3$ and simplify.

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Step 3.1

Divide each term in $3 x = π$ by $3$ .

$\frac{3 x}{3} = \frac{π}{3}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $3$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{π}{3}$

Step 3.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{3}$

Step 4

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.

$3 x = 2 π - π$

Step 5

Solve for $x$ .

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Step 5.1

Subtract $π$ from $2 π$ .

$3 x = π$

Step 5.2

Divide each term in $3 x = π$ by $3$ and simplify.

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Step 5.2.1

Divide each term in $3 x = π$ by $3$ .

$\frac{3 x}{3} = \frac{π}{3}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $3$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{π}{3}$

Step 5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{3}$

Step 6

Find the period of $\sec (3 x)$ .

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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 $3$ in the formula for period.

$\frac{2 π}{| 3 |}$

Step 6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$\frac{2 π}{3}$

Step 7

The period of the $\sec (3 x)$ function is $\frac{2 π}{3}$ so values will repeat every $\frac{2 π}{3}$ radians in both directions.

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