Trigonometry Examples

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

Step 1

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

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

Step 2

Simplify each term.

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Step 2.1

Apply the distributive property.

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

Step 2.2

Multiply $- 1$ by $1$ .

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

Step 3

Reorder the polynomial.

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

Step 4

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

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

Step 5

Factor the left side of the equation.

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Step 5.1

Factor $- 1$ out of $- u^{2} + 2 u - 1$ .

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Step 5.1.1

Factor $- 1$ out of $- u^{2}$ .

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

Step 5.1.2

Factor $- 1$ out of $2 u$ .

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

Step 5.1.3

Rewrite $- 1$ as $- 1 (1)$ .

$- u^{2} - (- 2 u) - 1 \cdot 1 = 0$

Step 5.1.4

Factor $- 1$ out of $- u^{2} - (- 2 u)$ .

$- (u^{2} - 2 u) - 1 \cdot 1 = 0$

Step 5.1.5

Factor $- 1$ out of $- (u^{2} - 2 u) - 1 (1)$ .

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

Step 5.2

Factor using the perfect square rule.

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Step 5.2.1

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

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

Step 5.2.2

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 5.2.3

Rewrite the polynomial.

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

Step 5.2.4

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 6

Divide each term in $- {(u - 1)}^{2} = 0$ by $- 1$ and simplify.

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Step 6.1

Divide each term in $- {(u - 1)}^{2} = 0$ by $- 1$ .

$\frac{- {(u - 1)}^{2}}{- 1} = \frac{0}{- 1}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Dividing two negative values results in a positive value.

$\frac{{(u - 1)}^{2}}{1} = \frac{0}{- 1}$

Step 6.2.2

Divide ${(u - 1)}^{2}$ by $1$ .

${(u - 1)}^{2} = \frac{0}{- 1}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Divide $0$ by $- 1$ .

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

Step 7

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

$u - 1 = 0$

Step 8

Add $1$ to both sides of the equation.

$u = 1$

Step 9

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

$\tan (x) = 1$

Step 10

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

$x = \arctan (1)$

Step 11

Simplify the right side.

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Step 11.1

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

$x = \frac{π}{4}$

Step 12

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 13

Simplify $π + \frac{π}{4}$ .

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Step 13.1

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 13.2

Combine fractions.

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Step 13.2.1

Combine $π$ and $\frac{4}{4}$ .

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

Step 13.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Step 13.3

Simplify the numerator.

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Step 13.3.1

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Step 13.3.2

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

Step 14

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

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Step 14.1

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

$\frac{π}{| b |}$

Step 14.2

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

$\frac{π}{| 1 |}$

Step 14.3

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

$\frac{π}{1}$

Step 14.4

Divide $π$ by $1$ .

$π$

Step 15

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 16

Consolidate the answers.

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