Solve for x in Radians tan(x)^2=1/3

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Trigonometry

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

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

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

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

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

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

Any root of $1$ is $1$ .

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

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

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

Combine and simplify the denominator.

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

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.

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

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

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

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

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

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

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

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

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

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

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

Simplify the right side.

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

$x = \frac{π}{6}$

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.

$x = π + \frac{π}{6}$

Simplify $π + \frac{π}{6}$ .

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

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

Combine fractions.

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Combine $π$ and $\frac{6}{6}$ .

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

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 + π}{6}$

Simplify the numerator.

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Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π + π}{6}$

Add $6 π$ and $π$ .

$x = \frac{7 π}{6}$

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.

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

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

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

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

Simplify the right side.

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

$x = - \frac{π}{6}$

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.

$x = - \frac{π}{6} - π$

Simplify the expression to find the second solution.

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Add $2 π$ to $- \frac{π}{6} - π$ .

$x = - \frac{π}{6} - π + 2 π$

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

$x = \frac{5 π}{6}$

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

$- \frac{π}{6} + π$

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

$π \cdot \frac{6}{6} - \frac{π}{6}$

Combine fractions.

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Combine $π$ and $\frac{6}{6}$ .

$\frac{π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$\frac{π \cdot 6 - π}{6}$

Simplify the numerator.

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Move $6$ to the left of $π$ .

$\frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$\frac{5 π}{6}$

List the new angles.

$x = \frac{5 π}{6}$

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 $n$

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 $n$

Consolidate the solutions.

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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 $n$

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 $n$

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