Calculus Examples

$\sec^{2} (x) + \tan (x) - 3 = 0$

Step 1

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

$(1 + \tan^{2} (x)) + \tan (x) - 3 = 0$

Step 2

Subtract $3$ from $1$ .

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

Step 3

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

${(u)}^{2} + u - 2 = 0$

Step 4

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

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Step 4.1

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 $1$ .

$- 1, 2$

Step 4.2

Write the factored form using these integers.

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

Step 5

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

$u - 1 = 0$

$u + 2 = 0$

Step 6

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

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Step 6.1

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

$u - 1 = 0$

Step 6.2

Add $1$ to both sides of the equation.

$u = 1$

Step 7

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

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Step 7.1

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

$u + 2 = 0$

Step 7.2

Subtract $2$ from both sides of the equation.

$u = - 2$

Step 8

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

$u = 1, - 2$

Step 9

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

$\tan (x) = 1, - 2$

Step 10

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

$\tan (x) = 1$

$\tan (x) = - 2$

Step 11

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

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Step 11.1

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

$x = \arctan (1)$

Step 11.2

Simplify the right side.

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Step 11.2.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$x = \frac{π}{4}$

Step 11.3

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = π + \frac{π}{4}$

Step 11.4

Simplify $π + \frac{π}{4}$ .

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Step 11.4.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 11.4.2

Combine fractions.

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Step 11.4.2.1

Combine $π$ and $\frac{4}{4}$ .

$x = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 11.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Step 11.4.3

Simplify the numerator.

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Step 11.4.3.1

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Step 11.4.3.2

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

Step 11.5

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

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Step 11.5.1

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

$\frac{π}{| b |}$

Step 11.5.2

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

$\frac{π}{| 1 |}$

Step 11.5.3

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

$\frac{π}{1}$

Step 11.5.4

Divide $π$ by $1$ .

$π$

Step 11.6

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

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

Step 12

Solve for $x$ in $\tan (x) = - 2$ .

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Step 12.1

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

$x = \arctan (- 2)$

Step 12.2

Simplify the right side.

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Step 12.2.1

Evaluate $\arctan (- 2)$ .

$x = - 1.10714871$

Step 12.3

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the third quadrant.

$x = - 1.10714871 - (3.14159265)$

Step 12.4

Simplify the expression to find the second solution.

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Step 12.4.1

Add $2 π$ to $- 1.10714871 - (3.14159265)$ .

$x = - 1.10714871 - (3.14159265) + 2 π$

Step 12.4.2

The resulting angle of $2.03444393$ is positive and coterminal with $- 1.10714871 - (3.14159265)$ .

$x = 2.03444393$

Step 12.5

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

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Step 12.5.1

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

$\frac{π}{| b |}$

Step 12.5.2

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

$\frac{π}{| 1 |}$

Step 12.5.3

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

$\frac{π}{1}$

Step 12.5.4

Divide $π$ by $1$ .

$π$

Step 12.6

Add $π$ to every negative angle to get positive angles.

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Step 12.6.1

Add $π$ to $- 1.10714871$ to find the positive angle.

$- 1.10714871 + π$

Step 12.6.2

Replace with decimal approximation.

$3.14159265 - 1.10714871$

Step 12.6.3

Subtract $1.10714871$ from $3.14159265$ .

$2.03444393$

Step 12.6.4

List the new angles.

$x = 2.03444393$

Step 12.7

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

$x = 2.03444393 + π n, 2.03444393 + π n$ , for any integer $n$

Step 13

List all of the solutions.

$x = \frac{π}{4} + π n, \frac{5 π}{4} + π n, 2.03444393 + π n, 2.03444393 + π n$ , for any integer $n$

Step 14

Consolidate the solutions.

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Step 14.1

Consolidate $\frac{π}{4} + π n$ and $\frac{5 π}{4} + π n$ to $\frac{π}{4} + π n$ .

$x = \frac{π}{4} + π n, 2.03444393 + π n, 2.03444393 + π n$ , for any integer $n$

Step 14.2

Consolidate $2.03444393 + π n$ and $2.03444393 + π n$ to $2.03444393 + π n$ .

$x = \frac{π}{4} + π n, 2.03444393 + π n$ , for any integer $n$