Solve for θ tan(theta)^2=3/8

$\tan^{2} (θ) = \frac{3}{8}$

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

$\tan (θ) = \pm \sqrt{\frac{3}{8}}$

Simplify $\pm \sqrt{\frac{3}{8}}$ .

Rewrite $\sqrt{\frac{3}{8}}$ as $\frac{\sqrt{3}}{\sqrt{8}}$ .

$\tan (θ) = \pm \frac{\sqrt{3}}{\sqrt{8}}$

Simplify the denominator.

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$\tan (θ) = \pm \frac{\sqrt{3}}{\sqrt{4 (2)}}$

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $\frac{\sqrt{3}}{2 \sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\tan (θ) = \pm \frac{\sqrt{3}}{2 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{3}}{2 \sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Move $\sqrt{2}$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

$\tan (θ) = \pm \frac{\sqrt{3} \sqrt{2}}{2 \cdot 2}$

Simplify the numerator.

Combine using the product rule for radicals.

$\tan (θ) = \pm \frac{\sqrt{3 \cdot 2}}{2 \cdot 2}$

Multiply $3$ by $2$ .

$\tan (θ) = \pm \frac{\sqrt{6}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$\tan (θ) = \pm \frac{\sqrt{6}}{4}$

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

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

$\tan (θ) = \frac{\sqrt{6}}{4}$

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

$\tan (θ) = - \frac{\sqrt{6}}{4}$

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

$\tan (θ) = \frac{\sqrt{6}}{4}, - \frac{\sqrt{6}}{4}$

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

$\tan (θ) = \frac{\sqrt{6}}{4}$

$\tan (θ) = - \frac{\sqrt{6}}{4}$

Solve for $θ$ in $\tan (θ) = \frac{\sqrt{6}}{4}$ .

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

$θ = \arctan (\frac{\sqrt{6}}{4})$

Simplify the right side.

Evaluate $\arctan (\frac{\sqrt{6}}{4})$ .

$θ = 0.54946724$

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.

$θ = (3.14159265) + 0.54946724$

Solve for $θ$ .

Remove parentheses.

$θ = 3.14159265 + 0.54946724$

Remove parentheses.

$θ = (3.14159265) + 0.54946724$

Add $3.14159265$ and $0.54946724$ .

$θ = 3.69105989$

Find the period of $\tan (θ)$ .

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 $\tan (θ)$ function is $π$ so values will repeat every $π$ radians in both directions.

$θ = 0.54946724 + π n, 3.69105989 + π n$ , for any integer $n$

Solve for $θ$ in $\tan (θ) = - \frac{\sqrt{6}}{4}$ .

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

$θ = \arctan (- \frac{\sqrt{6}}{4})$

Simplify the right side.

Evaluate $\arctan (- \frac{\sqrt{6}}{4})$ .

$θ = - 0.54946724$

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.

$θ = - 0.54946724 - (3.14159265)$

Simplify the expression to find the second solution.

Add $2 π$ to $- 0.54946724 - (3.14159265)$ .

$θ = - 0.54946724 - (3.14159265) + 2 π$

The resulting angle of $2.5921254$ is positive and coterminal with $- 0.54946724 - (3.14159265)$ .

$θ = 2.5921254$

Find the period of $\tan (θ)$ .

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.

Add $π$ to $- 0.54946724$ to find the positive angle.

$- 0.54946724 + π$

Replace with decimal approximation.

$3.14159265 - 0.54946724$

Subtract $0.54946724$ from $3.14159265$ .

$2.5921254$

List the new angles.

$θ = 2.5921254$

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

$θ = 2.5921254 + π n, 2.5921254 + π n$ , for any integer $n$

List all of the solutions.

$θ = 0.54946724 + π n, 3.69105989 + π n, 2.5921254 + π n, 2.5921254 + π n$ , for any integer $n$

Consolidate the solutions.

Consolidate $0.54946724 + π n$ and $3.69105989 + π n$ to $0.54946724 + π n$ .

$θ = 0.54946724 + π n, 2.5921254 + π n, 2.5921254 + π n$ , for any integer $n$

Consolidate $2.5921254 + π n$ and $2.5921254 + π n$ to $2.5921254 + π n$ .

$θ = 0.54946724 + π n, 2.5921254 + π n$ , for any integer $n$

Solve for θ tan(theta)^2=3/8

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