Solve for ? sec(x)^2-4=0

$\sec^{2} (x) - 4 = 0$

Add $4$ to both sides of the equation.

$\sec^{2} (x) = 4$

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

$\sec (x) = \pm \sqrt{4}$

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

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Simplify the right side of the equation.

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Rewrite $4$ as $2^{2}$ .

$\sec (x) = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$\sec (x) = \pm 2$

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

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

$\sec (x) = 2$

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

$\sec (x) = - 2$

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

$\sec (x) = 2, - 2$

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

$\sec (x) = 2$

$\sec (x) = - 2$

Set up the equation to solve for $x$ .

$\sec (x) = 2$

Solve the equation for $x$ .

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

$x = arcsec (2)$

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

$x = \frac{π}{3}$

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.

$x = 2 π - \frac{π}{3}$

Simplify $2 π - \frac{π}{3}$ .

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To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = \frac{2 π}{1} \cdot \frac{3}{3} - \frac{π}{3}$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{2 π \cdot 3}{1 \cdot 3} - \frac{π}{3}$

Multiply $3$ by $1$ .

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

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Simplify the numerator.

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Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Find the period.

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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.

$\frac{2 π}{| 1 |}$

Solve the equation.

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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 $\sec (x)$ 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 $n$

Set up the equation to solve for $x$ .

$\sec (x) = - 2$

Solve the equation for $x$ .

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

$x = arcsec (- 2)$

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

$x = \frac{2 π}{3}$

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.

$x = 2 π - \frac{2 π}{3}$

Simplify $2 π - \frac{2 π}{3}$ .

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To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = \frac{2 π}{1} \cdot \frac{3}{3} - \frac{2 π}{3}$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{2 π \cdot 3}{1 \cdot 3} - \frac{2 π}{3}$

Multiply $3$ by $1$ .

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

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Simplify the numerator.

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Multiply $3$ by $2$ .

$x = \frac{6 π - 2 π}{3}$

Subtract $2 π$ from $6 π$ .

$x = \frac{4 π}{3}$

Find the period.

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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.

$\frac{2 π}{| 1 |}$

Solve the equation.

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

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$

List all of the results found in the previous steps.

$x = \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

The complete solution is the set of all solutions.

$x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n, \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$

Consolidate the answers.

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Consolidate $\frac{π}{3} + 2 π n$ and $\frac{4 π}{3} + 2 π n$ to $\frac{π}{3} + π n$ .

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

Consolidate $\frac{5 π}{3} + 2 π n$ and $\frac{2 π}{3} + 2 π n$ to $\frac{2 π}{3} + π n$ .

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

Solve for ? sec(x)^2-4=0

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