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

$\tan (θ) = - \sqrt{3}$

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

$θ = \arctan (- \sqrt{3})$

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

$θ = - \frac{π}{3}$

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.

$θ = - \frac{π}{3} - π$

Simplify the expression to find the second solution.

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Simplify $- \frac{π}{3} - π$ .

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To write $- \frac{- π}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = - \frac{π}{3} + \frac{- π}{1} \cdot \frac{3}{3}$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$θ = - \frac{π}{3} + \frac{- π \cdot 3}{1 \cdot 3}$

Multiply $3$ by $1$ .

$θ = - \frac{π}{3} + \frac{- π \cdot 3}{3}$

Combine the numerators over the common denominator.

$θ = \frac{- π - π \cdot 3}{3}$

Simplify the numerator.

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Multiply $3$ by $- 1$ .

$θ = \frac{- π - 3 π}{3}$

Subtract $3 π$ from $- π$ .

$θ = \frac{- 4 π}{3}$

Move the negative in front of the fraction.

$θ = - \frac{4 π}{3}$

Add $2 π$ to $- \frac{4 π}{3}$ .

$θ = - \frac{4 π}{3} + 2 π$

The resulting angle of $\frac{2 π}{3}$ is positive and coterminal with $- \frac{4 π}{3}$ .

$θ = \frac{2 π}{3}$

Find the period.

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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.

$\frac{π}{| 1 |}$

Solve the equation.

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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 $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{π}{1} \cdot \frac{3}{3} - \frac{π}{3}$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$\frac{π \cdot 3}{1 \cdot 3} - \frac{π}{3}$

Multiply $3$ by $1$ .

$\frac{π \cdot 3}{3} - \frac{π}{3}$

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.

$θ = \frac{2 π}{3}$

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

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

Consolidate the answers.

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

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

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