Trigonometry Examples

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

Step 1

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

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

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

Step 2

Set $3 \tan^{2} (x) - 1$ equal to $0$ and solve for $x$ .

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Step 2.1

Set $3 \tan^{2} (x) - 1$ equal to $0$ .

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

Step 2.2

Solve $3 \tan^{2} (x) - 1 = 0$ for $x$ .

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Step 2.2.1

Add $1$ to both sides of the equation.

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

Step 2.2.2

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

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Step 2.2.2.1

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

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

Step 2.2.2.2

Simplify the left side.

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Step 2.2.2.2.1

Cancel the common factor of $3$ .

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Step 2.2.2.2.1.1

Cancel the common factor.

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

Step 2.2.2.2.1.2

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

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

Step 2.2.3

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

$\tan (x) = \pm \sqrt{\frac{1}{3}}$

Step 2.2.4

Simplify $\pm \sqrt{\frac{1}{3}}$ .

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Step 2.2.4.1

Rewrite $\sqrt{\frac{1}{3}}$ as $\frac{\sqrt{1}}{\sqrt{3}}$ .

$\tan (x) = \pm \frac{\sqrt{1}}{\sqrt{3}}$

Step 2.2.4.2

Any root of $1$ is $1$ .

$\tan (x) = \pm \frac{1}{\sqrt{3}}$

Step 2.2.4.3

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\tan (x) = \pm \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$

Step 2.2.4.4

Combine and simplify the denominator.

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Step 2.2.4.4.1

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

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

Step 2.2.4.4.2

Raise $\sqrt{3}$ to the power of $1$ .

$\tan (x) = \pm \frac{\sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Step 2.2.4.4.3

Raise $\sqrt{3}$ to the power of $1$ .

$\tan (x) = \pm \frac{\sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

Step 2.2.4.4.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\tan (x) = \pm \frac{\sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Step 2.2.4.4.5

Add $1$ and $1$ .

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

Step 2.2.4.4.6

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 2.2.4.4.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

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

Step 2.2.4.4.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\tan (x) = \pm \frac{\sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Step 2.2.4.4.6.3

Combine $\frac{1}{2}$ and $2$ .

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

Step 2.2.4.4.6.4

Cancel the common factor of $2$ .

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Step 2.2.4.4.6.4.1

Cancel the common factor.

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

Step 2.2.4.4.6.4.2

Rewrite the expression.

$\tan (x) = \pm \frac{\sqrt{3}}{3^{1}}$

Step 2.2.4.4.6.5

Evaluate the exponent.

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

Step 2.2.5

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

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Step 2.2.5.1

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

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

Step 2.2.5.2

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

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

Step 2.2.5.3

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

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

Step 2.2.6

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

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

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

Step 2.2.7

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

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Step 2.2.7.1

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

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

Step 2.2.7.2

Simplify the right side.

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Step 2.2.7.2.1

The exact value of $\arctan (\frac{\sqrt{3}}{3})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

Step 2.2.7.3

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.

$x = π + \frac{π}{6}$

Step 2.2.7.4

Simplify $π + \frac{π}{6}$ .

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Step 2.2.7.4.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} + \frac{π}{6}$

Step 2.2.7.4.2

Combine fractions.

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Step 2.2.7.4.2.1

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} + \frac{π}{6}$

Step 2.2.7.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 + π}{6}$

Step 2.2.7.4.3

Simplify the numerator.

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Step 2.2.7.4.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π + π}{6}$

Step 2.2.7.4.3.2

Add $6 π$ and $π$ .

$x = \frac{7 π}{6}$

Step 2.2.7.5

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

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Step 2.2.7.5.1

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

$\frac{π}{| b |}$

Step 2.2.7.5.2

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

$\frac{π}{| 1 |}$

Step 2.2.7.5.3

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

$\frac{π}{1}$

Step 2.2.7.5.4

Divide $π$ by $1$ .

$π$

Step 2.2.7.6

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

$x = \frac{π}{6} + π n, \frac{7 π}{6} + π n$ , for any integer $n$

Step 2.2.8

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

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Step 2.2.8.1

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

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

Step 2.2.8.2

Simplify the right side.

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Step 2.2.8.2.1

The exact value of $\arctan (- \frac{\sqrt{3}}{3})$ is $- \frac{π}{6}$ .

$x = - \frac{π}{6}$

Step 2.2.8.3

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.

$x = - \frac{π}{6} - π$

Step 2.2.8.4

Simplify the expression to find the second solution.

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Step 2.2.8.4.1

Add $2 π$ to $- \frac{π}{6} - π$ .

$x = - \frac{π}{6} - π + 2 π$

Step 2.2.8.4.2

The resulting angle of $\frac{5 π}{6}$ is positive and coterminal with $- \frac{π}{6} - π$ .

$x = \frac{5 π}{6}$

Step 2.2.8.5

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

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Step 2.2.8.5.1

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

$\frac{π}{| b |}$

Step 2.2.8.5.2

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

$\frac{π}{| 1 |}$

Step 2.2.8.5.3

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

$\frac{π}{1}$

Step 2.2.8.5.4

Divide $π$ by $1$ .

$π$

Step 2.2.8.6

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

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Step 2.2.8.6.1

Add $π$ to $- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + π$

Step 2.2.8.6.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$π \cdot \frac{6}{6} - \frac{π}{6}$

Step 2.2.8.6.3

Combine fractions.

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Step 2.2.8.6.3.1

Combine $π$ and $\frac{6}{6}$ .

$\frac{π \cdot 6}{6} - \frac{π}{6}$

Step 2.2.8.6.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 6 - π}{6}$

Step 2.2.8.6.4

Simplify the numerator.

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Step 2.2.8.6.4.1

Move $6$ to the left of $π$ .

$\frac{6 \cdot π - π}{6}$

Step 2.2.8.6.4.2

Subtract $π$ from $6 π$ .

$\frac{5 π}{6}$

Step 2.2.8.6.5

List the new angles.

$x = \frac{5 π}{6}$

Step 2.2.8.7

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

$x = \frac{5 π}{6} + π n, \frac{5 π}{6} + π n$ , for any integer $n$

Step 2.2.9

List all of the solutions.

$x = \frac{π}{6} + π n, \frac{7 π}{6} + π n, \frac{5 π}{6} + π n, \frac{5 π}{6} + π n$ , for any integer $n$

Step 2.2.10

Consolidate the solutions.

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Step 2.2.10.1

Consolidate $\frac{π}{6} + π n$ and $\frac{7 π}{6} + π n$ to $\frac{π}{6} + π n$ .

$x = \frac{π}{6} + π n, \frac{5 π}{6} + π n, \frac{5 π}{6} + π n$ , for any integer $n$

Step 2.2.10.2

Consolidate $\frac{5 π}{6} + π n$ and $\frac{5 π}{6} + π n$ to $\frac{5 π}{6} + π n$ .

$x = \frac{π}{6} + π n, \frac{5 π}{6} + π n$ , for any integer $n$

Step 3

Set $\tan^{2} (x) - 3$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $\tan^{2} (x) - 3$ equal to $0$ .

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

Step 3.2

Solve $\tan^{2} (x) - 3 = 0$ for $x$ .

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Step 3.2.1

Add $3$ to both sides of the equation.

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

Step 3.2.2

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

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

Step 3.2.3

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

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Step 3.2.3.1

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

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

Step 3.2.3.2

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

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

Step 3.2.3.3

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

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

Step 3.2.4

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

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

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

Step 3.2.5

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

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Step 3.2.5.1

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

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

Step 3.2.5.2

Simplify the right side.

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Step 3.2.5.2.1

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

$x = \frac{π}{3}$

Step 3.2.5.3

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 3.2.5.4

Simplify $π + \frac{π}{3}$ .

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Step 3.2.5.4.1

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 3.2.5.4.2

Combine fractions.

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Step 3.2.5.4.2.1

Combine $π$ and $\frac{3}{3}$ .

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

Step 3.2.5.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 3 + π}{3}$

Step 3.2.5.4.3

Simplify the numerator.

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Step 3.2.5.4.3.1

Move $3$ to the left of $π$ .

$x = \frac{3 \cdot π + π}{3}$

Step 3.2.5.4.3.2

Add $3 π$ and $π$ .

$x = \frac{4 π}{3}$

Step 3.2.5.5

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

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Step 3.2.5.5.1

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

$\frac{π}{| b |}$

Step 3.2.5.5.2

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

$\frac{π}{| 1 |}$

Step 3.2.5.5.3

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

$\frac{π}{1}$

Step 3.2.5.5.4

Divide $π$ by $1$ .

$π$

Step 3.2.5.6

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 3.2.6

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

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Step 3.2.6.1

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

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

Step 3.2.6.2

Simplify the right side.

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Step 3.2.6.2.1

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

$x = - \frac{π}{3}$

Step 3.2.6.3

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 3.2.6.4

Simplify the expression to find the second solution.

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Step 3.2.6.4.1

Add $2 π$ to $- \frac{π}{3} - π$ .

$x = - \frac{π}{3} - π + 2 π$

Step 3.2.6.4.2

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

$x = \frac{2 π}{3}$

Step 3.2.6.5

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

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Step 3.2.6.5.1

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

$\frac{π}{| b |}$

Step 3.2.6.5.2

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

$\frac{π}{| 1 |}$

Step 3.2.6.5.3

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

$\frac{π}{1}$

Step 3.2.6.5.4

Divide $π$ by $1$ .

$π$

Step 3.2.6.6

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

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Step 3.2.6.6.1

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

$- \frac{π}{3} + π$

Step 3.2.6.6.2

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

$π \cdot \frac{3}{3} - \frac{π}{3}$

Step 3.2.6.6.3

Combine fractions.

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Step 3.2.6.6.3.1

Combine $π$ and $\frac{3}{3}$ .

$\frac{π \cdot 3}{3} - \frac{π}{3}$

Step 3.2.6.6.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 3 - π}{3}$

Step 3.2.6.6.4

Simplify the numerator.

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Step 3.2.6.6.4.1

Move $3$ to the left of $π$ .

$\frac{3 \cdot π - π}{3}$

Step 3.2.6.6.4.2

Subtract $π$ from $3 π$ .

$\frac{2 π}{3}$

Step 3.2.6.6.5

List the new angles.

$x = \frac{2 π}{3}$

Step 3.2.6.7

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 3.2.7

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 3.2.8

Consolidate the solutions.

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Step 3.2.8.1

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 3.2.8.2

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 $n$

Step 4

The final solution is all the values that make $(3 \tan^{2} (x) - 1) (\tan^{2} (x) - 3) = 0$ true.

$x = \frac{π}{6} + π n, \frac{5 π}{6} + π n, \frac{π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer $n$

Step 5

Consolidate the answers.

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Step 5.1

Consolidate $\frac{π}{6} + π n$ and $\frac{2 π}{3} + π n$ to $\frac{π}{6} + \frac{π n}{2}$ .

$x = \frac{π}{6} + \frac{π n}{2}, \frac{5 π}{6} + π n, \frac{π}{3} + π n$ , for any integer $n$

Step 5.2

Consolidate $\frac{5 π}{6} + π n$ and $\frac{π}{3} + π n$ to $\frac{π}{3} + \frac{π n}{2}$ .

$x = \frac{π}{6} + \frac{π n}{2}, \frac{π}{3} + \frac{π n}{2}$ , for any integer $n$