Trigonometry Examples

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

Step 1

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

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

Step 2

Subtract $2 \tan^{2} (x)$ from $\tan^{2} (x)$ .

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

Step 3

Reorder the polynomial.

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

Step 4

Move all terms not containing $\tan (x)$ to the right side of the equation.

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Step 4.1

Subtract $1$ from both sides of the equation.

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

Step 4.2

Subtract $1$ from $- 2$ .

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

Step 5

Divide each term in $- \tan^{2} (x) = - 3$ by $- 1$ and simplify.

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Step 5.1

Divide each term in $- \tan^{2} (x) = - 3$ by $- 1$ .

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

Step 5.2

Simplify the left side.

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Step 5.2.1

Dividing two negative values results in a positive value.

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

Step 5.2.2

Divide $\tan^{2} (x)$ by $1$ .

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

Step 5.3

Simplify the right side.

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Step 5.3.1

Divide $- 3$ by $- 1$ .

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

Step 6

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

$\tan (x) = \pm \sqrt{3}$

Step 7

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

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Step 7.1

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

$\tan (x) = \sqrt{3}$

Step 7.2

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

$\tan (x) = - \sqrt{3}$

Step 7.3

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

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

Step 8

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

$\tan (x) = \sqrt{3}$

$\tan (x) = - \sqrt{3}$

Step 9

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

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Step 9.1

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

$x = \arctan (\sqrt{3})$

Step 9.2

Simplify the right side.

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Step 9.2.1

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

$x = \frac{π}{3}$

Step 9.3

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.

$x = π + \frac{π}{3}$

Step 9.4

Simplify $π + \frac{π}{3}$ .

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Step 9.4.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = π \cdot \frac{3}{3} + \frac{π}{3}$

Step 9.4.2

Combine fractions.

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Step 9.4.2.1

Combine $π$ and $\frac{3}{3}$ .

$x = \frac{π \cdot 3}{3} + \frac{π}{3}$

Step 9.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 3 + π}{3}$

Step 9.4.3

Simplify the numerator.

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Step 9.4.3.1

Move $3$ to the left of $π$ .

$x = \frac{3 \cdot π + π}{3}$

Step 9.4.3.2

Add $3 π$ and $π$ .

$x = \frac{4 π}{3}$

Step 9.5

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

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Step 9.5.1

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

$\frac{π}{| b |}$

Step 9.5.2

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

$\frac{π}{| 1 |}$

Step 9.5.3

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

$\frac{π}{1}$

Step 9.5.4

Divide $π$ by $1$ .

$π$

Step 9.6

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

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

Step 10

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

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Step 10.1

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

$x = \arctan (- \sqrt{3})$

Step 10.2

Simplify the right side.

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Step 10.2.1

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

$x = - \frac{π}{3}$

Step 10.3

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.

$x = - \frac{π}{3} - π$

Step 10.4

Simplify the expression to find the second solution.

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Step 10.4.1

Add $2 π$ to $- \frac{π}{3} - π$ .

$x = - \frac{π}{3} - π + 2 π$

Step 10.4.2

The resulting angle of $\frac{2 π}{3}$ is positive and coterminal with $- \frac{π}{3} - π$ .

$x = \frac{2 π}{3}$

Step 10.5

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

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Step 10.5.1

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

$\frac{π}{| b |}$

Step 10.5.2

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

$\frac{π}{| 1 |}$

Step 10.5.3

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

$\frac{π}{1}$

Step 10.5.4

Divide $π$ by $1$ .

$π$

Step 10.6

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

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Step 10.6.1

Add $π$ to $- \frac{π}{3}$ to find the positive angle.

$- \frac{π}{3} + π$

Step 10.6.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$π \cdot \frac{3}{3} - \frac{π}{3}$

Step 10.6.3

Combine fractions.

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Step 10.6.3.1

Combine $π$ and $\frac{3}{3}$ .

$\frac{π \cdot 3}{3} - \frac{π}{3}$

Step 10.6.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 3 - π}{3}$

Step 10.6.4

Simplify the numerator.

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Step 10.6.4.1

Move $3$ to the left of $π$ .

$\frac{3 \cdot π - π}{3}$

Step 10.6.4.2

Subtract $π$ from $3 π$ .

$\frac{2 π}{3}$

Step 10.6.5

List the new angles.

$x = \frac{2 π}{3}$

Step 10.7

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

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

Step 11

List all of the solutions.

$x = \frac{π}{3} + π n, \frac{4 π}{3} + π n, \frac{2 π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer $n$

Step 12

Consolidate the solutions.

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Step 12.1

Consolidate $\frac{π}{3} + π n$ and $\frac{4 π}{3} + π n$ to $\frac{π}{3} + π n$ .

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

Step 12.2

Consolidate $\frac{2 π}{3} + π n$ and $\frac{2 π}{3} + π n$ to $\frac{2 π}{3} + π n$ .

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