Evaluate 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.

$θ = \arctan (\sqrt{3})$

Simplify the right side.

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

$θ = \frac{π}{3}$

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.

$θ = π + \frac{π}{3}$

Simplify $π + \frac{π}{3}$ .

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

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

Combine fractions.

Combine $π$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 3 + π}{3}$

Simplify the numerator.

Move $3$ to the left of $π$ .

$θ = \frac{3 \cdot π + π}{3}$

Add $3 π$ and $π$ .

$θ = \frac{4 π}{3}$

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.

$θ = \frac{π}{3} + π n, \frac{4 π}{3} + π n$ , for any integer $n$

Consolidate the answers.

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

Evaluate tan(theta) = square root of 3

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