Solve for x in Radians tan(x)^5-9tan(x)=0

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Trigonometry

$\tan^{5} (x) - 9 \tan (x) = 0$

Factor $\tan^{5} (x) - 9 \tan (x)$ .

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Factor $\tan (x)$ out of $\tan^{5} (x) - 9 \tan (x)$ .

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Factor $\tan (x)$ out of $\tan^{5} (x)$ .

$\tan (x) \tan^{4} (x) - 9 \tan (x) = 0$

Factor $\tan (x)$ out of $- 9 \tan (x)$ .

$\tan (x) \tan^{4} (x) + \tan (x) \cdot - 9 = 0$

Factor $\tan (x)$ out of $\tan (x) \tan^{4} (x) + \tan (x) \cdot - 9$ .

$\tan (x) (\tan^{4} (x) - 9) = 0$

Rewrite $\tan^{4} (x)$ as ${(\tan^{2} (x))}^{2}$ .

$\tan (x) ({(\tan^{2} (x))}^{2} - 9) = 0$

Rewrite $9$ as $3^{2}$ .

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

Factor.

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Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \tan^{2} (x)$ and $b = 3$ .

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

Remove unnecessary parentheses.

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

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

$\tan (x) = 0$

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

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

Set $\tan (x)$ equal to $0$ and solve for $x$ .

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Set $\tan (x)$ equal to $0$ .

$\tan (x) = 0$

Solve $\tan (x) = 0$ for $x$ .

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

$x = \arctan (0)$

Simplify the right side.

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The exact value of $\arctan (0)$ is $0$ .

$x = 0$

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.

$x = π + 0$

Add $π$ and $0$ .

$x = π$

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

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

$x = π n, π + π n$ , for any integer $n$

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

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

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

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

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Subtract $3$ from both sides of the equation.

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

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

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

Simplify $\pm \sqrt{- 3}$ .

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

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

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The inverse tangent of $\arctan (i \sqrt{3})$ is undefined.

Undefined

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

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

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

The inverse tangent of $\arctan (- i \sqrt{3})$ is undefined.

Undefined

List all of the solutions.

No solution

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

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

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

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

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Add $3$ to both sides of the equation.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify the right side.

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

$x = \frac{π}{3}$

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.

$x = π + \frac{π}{3}$

Simplify $π + \frac{π}{3}$ .

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

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

Combine fractions.

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Combine $π$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

$x = \frac{π \cdot 3 + π}{3}$

Simplify the numerator.

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Move $3$ to the left of $π$ .

$x = \frac{3 \cdot π + π}{3}$

Add $3 π$ and $π$ .

$x = \frac{4 π}{3}$

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

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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 $n$

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

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

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

Simplify the right side.

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

$x = - \frac{π}{3}$

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.

$x = - \frac{π}{3} - π$

Simplify the expression to find the second solution.

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Add $2 π$ to $- \frac{π}{3} - π$ .

$x = - \frac{π}{3} - π + 2 π$

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

$x = \frac{2 π}{3}$

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

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

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

$- \frac{π}{3} + π$

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

$π \cdot \frac{3}{3} - \frac{π}{3}$

Combine fractions.

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Combine $π$ and $\frac{3}{3}$ .

$\frac{π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$\frac{π \cdot 3 - π}{3}$

Simplify the numerator.

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Move $3$ to the left of $π$ .

$\frac{3 \cdot π - π}{3}$

Subtract $π$ from $3 π$ .

$\frac{2 π}{3}$

List the new angles.

$x = \frac{2 π}{3}$

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 $n$

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 $n$

Consolidate the solutions.

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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 $n$

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 $n$

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

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

Consolidate the answers.

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

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