Trigonometry Examples

$2 \sec^{2} (θ) - \tan^{4} (θ) = - 1$

Step 1

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

$2 (1 + \tan^{2} (θ)) - \tan^{4} (θ) = - 1$

Step 2

Simplify each term.

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Step 2.1

Apply the distributive property.

$2 \cdot 1 + 2 \tan^{2} (θ) - \tan^{4} (θ) = - 1$

Step 2.2

Multiply $2$ by $1$ .

$2 + 2 \tan^{2} (θ) - \tan^{4} (θ) = - 1$

Step 3

Reorder the polynomial.

$- \tan^{4} (θ) + 2 \tan^{2} (θ) + 2 = - 1$

Step 4

Substitute $u = \tan^{2} (θ)$ into the equation. This will make the quadratic formula easy to use.

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

$u = \tan^{2} (θ)$

Step 5

Add $1$ to both sides of the equation.

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

Step 6

Add $2$ and $1$ .

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

Step 7

Factor the left side of the equation.

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Step 7.1

Factor $- 1$ out of $- u^{2} + 2 u + 3$ .

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Step 7.1.1

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

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

Step 7.1.2

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

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

Step 7.1.3

Rewrite $3$ as $- 1 (- 3)$ .

$- (u^{2}) - (- 2 u) - 1 \cdot - 3 = 0$

Step 7.1.4

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

$- (u^{2} - 2 u) - 1 \cdot - 3 = 0$

Step 7.1.5

Factor $- 1$ out of $- (u^{2} - 2 u) - 1 (- 3)$ .

$- (u^{2} - 2 u - 3) = 0$

Step 7.2

Factor.

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Step 7.2.1

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

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Step 7.2.1.1

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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 7.2.1.2

Write the factored form using these integers.

$- ((u - 3) (u + 1)) = 0$

Step 7.2.2

Remove unnecessary parentheses.

$- (u - 3) (u + 1) = 0$

Step 8

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

$u - 3 = 0$

$u + 1 = 0$

Step 9

Set $u - 3$ equal to $0$ and solve for $u$ .

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Step 9.1

Set $u - 3$ equal to $0$ .

$u - 3 = 0$

Step 9.2

Add $3$ to both sides of the equation.

$u = 3$

Step 10

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

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Step 10.1

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

$u + 1 = 0$

Step 10.2

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 11

The final solution is all the values that make $- (u - 3) (u + 1) = 0$ true.

$u = 3, - 1$

Step 12

Substitute the real value of $u = \tan^{2} (θ)$ back into the solved equation.

$\tan^{2} (θ) = 3$

${(\tan^{2} (θ))}^{1} = - 1$

Step 13

Solve the first equation for $\tan (θ)$ .

$\tan^{2} (θ) = 3$

Step 14

Solve the equation for $\tan (θ)$ .

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Step 14.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\tan (θ) = \pm \sqrt{3}$

Step 14.2

The complete solution is the result of both the positive and negative portions of the solution.

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Step 14.2.1

First, use the positive value of the $\pm$ to find the first solution.

$\tan (θ) = \sqrt{3}$

Step 14.2.2

Next, use the negative value of the $\pm$ to find the second solution.

$\tan (θ) = - \sqrt{3}$

Step 14.2.3

The complete solution is the result of both the positive and negative portions of the solution.

$\tan (θ) = \sqrt{3}, - \sqrt{3}$

Step 15

Solve the second equation for $\tan (θ)$ .

${(\tan^{2} (θ))}^{1} = - 1$

Step 16

Solve the equation for $\tan (θ)$ .

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Step 16.1

Remove parentheses.

$\tan^{2} (θ) = - 1$

Step 16.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\tan (θ) = \pm \sqrt{- 1}$

Step 16.3

Rewrite $\sqrt{- 1}$ as $i$ .

$\tan (θ) = \pm i$

Step 16.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 16.4.1

First, use the positive value of the $\pm$ to find the first solution.

$\tan (θ) = i$

Step 16.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$\tan (θ) = - i$

Step 16.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$\tan (θ) = i, - i$

Step 17

The solution to $- \tan^{4} (θ) + 2 \tan^{2} (θ) + 2 = - 1$ is $\tan (θ) = \sqrt{3}, - \sqrt{3}, i, - i$ .

$\tan (θ) = \sqrt{3}, - \sqrt{3}, i, - i$

Step 18

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

$\tan (θ) = \sqrt{3}$

$\tan (θ) = - \sqrt{3}$

$\tan (θ) = i$

$\tan (θ) = - i$

Step 19

Solve for $θ$ in $\tan (θ) = \sqrt{3}$ .

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Step 19.1

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

$θ = \arctan (\sqrt{3})$

Step 19.2

Simplify the right side.

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Step 19.2.1

The exact value of $\arctan (\sqrt{3})$ is $60$ .

$θ = 60$

Step 19.3

The tangent function is positive in the first and third quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the fourth quadrant.

$θ = 180 + 60$

Step 19.4

Add $180$ and $60$ .

$θ = 240$

Step 19.5

Find the period of $\tan (θ)$ .

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Step 19.5.1

The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

Step 19.5.2

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

$\frac{180}{| 1 |}$

Step 19.5.3

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

$\frac{180}{1}$

Step 19.5.4

Divide $180$ by $1$ .

$180$

Step 19.6

The period of the $\tan (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 60 + 180 n, 240 + 180 n$ , for any integer $n$

Step 20

Solve for $θ$ in $\tan (θ) = - \sqrt{3}$ .

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Step 20.1

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

$θ = \arctan (- \sqrt{3})$

Step 20.2

Simplify the right side.

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Step 20.2.1

The exact value of $\arctan (- \sqrt{3})$ is $- 60$ .

$θ = - 60$

Step 20.3

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the third quadrant.

$θ = - 60 - 180$

Step 20.4

Simplify the expression to find the second solution.

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Step 20.4.1

Add $360 °$ to $- 60 - 180 °$ .

$θ = - 60 - 180 ° + 360 °$

Step 20.4.2

The resulting angle of $120 °$ is positive and coterminal with $- 60 - 180$ .

$θ = 120 °$

Step 20.5

Find the period of $\tan (θ)$ .

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Step 20.5.1

The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

Step 20.5.2

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

$\frac{180}{| 1 |}$

Step 20.5.3

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

$\frac{180}{1}$

Step 20.5.4

Divide $180$ by $1$ .

$180$

Step 20.6

Add $180$ to every negative angle to get positive angles.

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Step 20.6.1

Add $180$ to $- 60$ to find the positive angle.

$- 60 + 180$

Step 20.6.2

Subtract $60$ from $180$ .

$120$

Step 20.6.3

List the new angles.

$θ = 120$

Step 20.7

The period of the $\tan (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 120 + 180 n, 120 + 180 n$ , for any integer $n$

Step 21

Solve for $θ$ in $\tan (θ) = i$ .

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Step 21.1

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

$θ = \arctan (i)$

Step 21.2

The inverse tangent of $\arctan (i)$ is undefined.

Undefined

Step 22

Solve for $θ$ in $\tan (θ) = - i$ .

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Step 22.1

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

$θ = \arctan (- i)$

Step 22.2

The inverse tangent of $\arctan (- i)$ is undefined.

Undefined

Step 23

List all of the solutions.

$θ = 60 + 180 n, 240 + 180 n, 120 + 180 n, 120 + 180 n$ , for any integer $n$

Step 24

Consolidate the solutions.

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Step 24.1

Consolidate $60 + 180 n$ and $240 + 180 n$ to $60 + 180 n$ .

$θ = 60 + 180 n, 120 + 180 n, 120 + 180 n$ , for any integer $n$

Step 24.2

Consolidate $120 + 180 n$ and $120 + 180 n$ to $120 + 180 n$ .

$θ = 60 + 180 n, 120 + 180 n$ , for any integer $n$