Solve for θ tan(theta)=-( square root of 3)/3

$\tan (θ) = - \frac{\sqrt{3}}{3}$

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

$θ = \arctan (- \frac{\sqrt{3}}{3})$

Simplify the right side.

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

$θ = - \frac{π}{6}$

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.

$θ = - \frac{π}{6} - π$

Simplify the expression to find the second solution.

Add $2 π$ to $- \frac{π}{6} - π$ .

$θ = - \frac{π}{6} - π + 2 π$

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

$θ = \frac{5 π}{6}$

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 $- \frac{π}{6}$ 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.

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.

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.

$θ = \frac{5 π}{6}$

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

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

Consolidate the answers.

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

Solve for θ tan(theta)=-( square root of 3)/3

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