Solve for x in Radians sec(x)^2-sec(x)=2

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

Subtract $2$ from both sides of the equation.

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

Factor $\sec^{2} (x) - \sec (x) - 2$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

$(\sec (x) - 2) (\sec (x) + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\sec (x) - 2 = 0$

$\sec (x) + 1 = 0$

Set $\sec (x) - 2$ equal to $0$ and solve for $x$ .

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Set $\sec (x) - 2$ equal to $0$ .

$\sec (x) - 2 = 0$

Solve $\sec (x) - 2 = 0$ for $x$ .

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Add $2$ to both sides of the equation.

$\sec (x) = 2$

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

$x = arcsec (2)$

Simplify the right side.

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

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

Combine fractions.

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Combine $2 π$ and $\frac{3}{3}$ .

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

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 of $\sec (x)$ .

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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.

$\frac{2 π}{| 1 |}$

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 $\sec (x) + 1$ equal to $0$ and solve for $x$ .

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Set $\sec (x) + 1$ equal to $0$ .

$\sec (x) + 1 = 0$

Solve $\sec (x) + 1 = 0$ for $x$ .

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Subtract $1$ from both sides of the equation.

$\sec (x) = - 1$

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

$x = arcsec (- 1)$

Simplify the right side.

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

$x = π$

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.

$x = 2 π - π$

Subtract $π$ from $2 π$ .

$x = π$

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

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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.

$\frac{2 π}{| 1 |}$

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.

$x = π + 2 π n$ , for any integer $n$

The final solution is all the values that make $(\sec (x) - 2) (\sec (x) + 1) = 0$ true.

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

Consolidate the answers.

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

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