Trigonometry Examples

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

Step 1

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

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

Step 2

Reorder the polynomial.

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

Step 3

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

${(u)}^{2} + 2 (u) + 1 = 0$

Step 4

Factor using the perfect square rule.

Tap for more steps...

Step 4.1

Rewrite $1$ as $1^{2}$ .

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

Step 4.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

Step 4.3

Rewrite the polynomial.

$u^{2} + 2 \cdot u \cdot 1 + 1^{2} = 0$

Step 4.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 1$ .

${(u + 1)}^{2} = 0$

Step 5

Set the $u + 1$ equal to $0$ .

$u + 1 = 0$

Step 6

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 7

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

$\tan (x) = - 1$

Step 8

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

$x = \arctan (- 1)$

Step 9

Simplify the right side.

Tap for more steps...

Step 9.1

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

$x = - \frac{π}{4}$

Step 10

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.

$x = - \frac{π}{4} - π$

Step 11

Simplify the expression to find the second solution.

Tap for more steps...

Step 11.1

Add $2 π$ to $- \frac{π}{4} - π$ .

$x = - \frac{π}{4} - π + 2 π$

Step 11.2

The resulting angle of $\frac{3 π}{4}$ is positive and coterminal with $- \frac{π}{4} - π$ .

$x = \frac{3 π}{4}$

Step 12

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

Tap for more steps...

Step 12.1

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

$\frac{π}{| b |}$

Step 12.2

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

$\frac{π}{| 1 |}$

Step 12.3

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

$\frac{π}{1}$

Step 12.4

Divide $π$ by $1$ .

$π$

Step 13

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

Tap for more steps...

Step 13.1

Add $π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + π$

Step 13.2

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

$π \cdot \frac{4}{4} - \frac{π}{4}$

Step 13.3

Combine fractions.

Tap for more steps...

Step 13.3.1

Combine $π$ and $\frac{4}{4}$ .

$\frac{π \cdot 4}{4} - \frac{π}{4}$

Step 13.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 4 - π}{4}$

Step 13.4

Simplify the numerator.

Tap for more steps...

Step 13.4.1

Move $4$ to the left of $π$ .

$\frac{4 \cdot π - π}{4}$

Step 13.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{4}$

Step 13.5

List the new angles.

$x = \frac{3 π}{4}$

Step 14

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

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

Step 15

Consolidate the answers.

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