Trigonometry Examples

$\tan^{2} (x) - 3 \sec (x) = - 3$

Step 1

Replace the $\tan^{2} (x)$ with $\sec^{2} (x) - 1$ based on the $\tan^{2} (x) + 1 = \sec^{2} (x)$ identity.

$(\sec^{2} (x) - 1) - 3 \sec (x) = - 3$

Step 2

Reorder the polynomial.

$\sec^{2} (x) - 3 \sec (x) - 1 = - 3$

Step 3

Substitute $u$ for $\sec (x)$ .

${(u)}^{2} - 3 u - 1 = - 3$

Step 4

Add $3$ to both sides of the equation.

$u^{2} - 3 u - 1 + 3 = 0$

Step 5

Add $- 1$ and $3$ .

$u^{2} - 3 u + 2 = 0$

Step 6

Factor $u^{2} - 3 u + 2$ using the AC method.

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Step 6.1

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 $- 3$ .

$- 2, - 1$

Step 6.2

Write the factored form using these integers.

$(u - 2) (u - 1) = 0$

Step 7

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

$u - 2 = 0$

$u - 1 = 0$

Step 8

Set $u - 2$ equal to $0$ and solve for $u$ .

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Step 8.1

Set $u - 2$ equal to $0$ .

$u - 2 = 0$

Step 8.2

Add $2$ to both sides of the equation.

$u = 2$

Step 9

Set $u - 1$ equal to $0$ and solve for $u$ .

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Step 9.1

Set $u - 1$ equal to $0$ .

$u - 1 = 0$

Step 9.2

Add $1$ to both sides of the equation.

$u = 1$

Step 10

The final solution is all the values that make $(u - 2) (u - 1) = 0$ true.

$u = 2, 1$

Step 11

Substitute $\sec (x)$ for $u$ .

$\sec (x) = 2, 1$

Step 12

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

$\sec (x) = 2$

$\sec (x) = 1$

Step 13

Solve for $x$ in $\sec (x) = 2$ .

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Step 13.1

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

$x = arcsec (2)$

Step 13.2

Simplify the right side.

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Step 13.2.1

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

$x = \frac{π}{3}$

Step 13.3

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.

$x = 2 π - \frac{π}{3}$

Step 13.4

Simplify $2 π - \frac{π}{3}$ .

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Step 13.4.1

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 13.4.2

Combine fractions.

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Step 13.4.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 13.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 13.4.3

Simplify the numerator.

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Step 13.4.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 13.4.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 13.5

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

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Step 13.5.1

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

$\frac{2 π}{| b |}$

Step 13.5.2

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

$\frac{2 π}{| 1 |}$

Step 13.5.3

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

$\frac{2 π}{1}$

Step 13.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 13.6

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 $n$

Step 14

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

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Step 14.1

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

$x = arcsec (1)$

Step 14.2

Simplify the right side.

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Step 14.2.1

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

$x = 0$

Step 14.3

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 14.4

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 14.5

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

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Step 14.5.1

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

$\frac{2 π}{| b |}$

Step 14.5.2

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

$\frac{2 π}{| 1 |}$

Step 14.5.3

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

$\frac{2 π}{1}$

Step 14.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 14.6

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 15

List all of the solutions.

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

Step 16

Consolidate $2 π n$ and $2 π + 2 π n$ to $2 π n$ .

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