Trigonometry Examples

$\sec^{2} (x) - 1 = 0$

Step 1

Add $1$ to both sides of the equation.

$\sec^{2} (x) = 1$

Step 2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\sec (x) = \pm \sqrt{1}$

Step 3

Any root of $1$ is $1$ .

$\sec (x) = \pm 1$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.1

First, use the positive value of the $\pm$ to find the first solution.

$\sec (x) = 1$

Step 4.2

Next, use the negative value of the $\pm$ to find the second solution.

$\sec (x) = - 1$

Step 4.3

The complete solution is the result of both the positive and negative portions of the solution.

$\sec (x) = 1, - 1$

Step 5

Set up each of the solutions to solve for $x$ .

$\sec (x) = 1$

$\sec (x) = - 1$

Step 6

Solve for $x$ in $\sec (x) = 1$ .

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Step 6.1

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

$x = arcsec (1)$

Step 6.2

Simplify the right side.

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Step 6.2.1

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

$x = 0$

Step 6.3

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.

$x = 2 π - 0$

Step 6.4

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 6.5

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

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Step 6.5.1

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

$\frac{2 π}{| b |}$

Step 6.5.2

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

$\frac{2 π}{| 1 |}$

Step 6.5.3

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

$\frac{2 π}{1}$

Step 6.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.6

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

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

Step 7

Solve for $x$ in $\sec (x) = - 1$ .

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Step 7.1

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

$x = arcsec (- 1)$

Step 7.2

Simplify the right side.

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Step 7.2.1

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

$x = π$

Step 7.3

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.

$x = 2 π - π$

Step 7.4

Subtract $π$ from $2 π$ .

$x = π$

Step 7.5

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

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Step 7.5.1

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

$\frac{2 π}{| b |}$

Step 7.5.2

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

$\frac{2 π}{| 1 |}$

Step 7.5.3

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

$\frac{2 π}{1}$

Step 7.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 7.6

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

$x = π + 2 π n$ , for any integer $n$

Step 8

List all of the solutions.

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

Step 9

Consolidate the answers.

$x = π n$ , for any integer $n$