Solve for θ tan(theta/2) = square root of 3

$\tan (\frac{θ}{2}) = \sqrt{3}$

Step 1

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

$\frac{θ}{2} = \arctan (\sqrt{3})$

Step 2

Simplify the right side.

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

$\frac{θ}{2} = \frac{π}{3}$

Step 3

Multiply both sides of the equation by $2$ .

$2 \frac{θ}{2} = 2 \frac{π}{3}$

Step 4

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{θ}{2} = 2 \frac{π}{3}$

Rewrite the expression.

$θ = 2 \frac{π}{3}$

Simplify the right side.

Combine $2$ and $\frac{π}{3}$ .

$θ = \frac{2 π}{3}$

Step 5

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.

$\frac{θ}{2} = π + \frac{π}{3}$

Step 6

Solve for $θ$ .

Multiply both sides of the equation by $2$ .

$2 \frac{θ}{2} = 2 (π + \frac{π}{3})$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{θ}{2} = 2 (π + \frac{π}{3})$

Rewrite the expression.

$θ = 2 (π + \frac{π}{3})$

Simplify the right side.

Simplify $2 (π + \frac{π}{3})$ .

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

$θ = 2 (π \cdot \frac{3}{3} + \frac{π}{3})$

Combine fractions.

Combine $π$ and $\frac{3}{3}$ .

$θ = 2 (\frac{π \cdot 3}{3} + \frac{π}{3})$

Combine the numerators over the common denominator.

$θ = 2 \frac{π \cdot 3 + π}{3}$

Simplify the numerator.

Move $3$ to the left of $π$ .

$θ = 2 \frac{3 \cdot π + π}{3}$

Add $3 π$ and $π$ .

$θ = 2 \frac{4 π}{3}$

Multiply $2 \frac{4 π}{3}$ .

Combine $2$ and $\frac{4 π}{3}$ .

$θ = \frac{2 (4 π)}{3}$

Multiply $4$ by $2$ .

$θ = \frac{8 π}{3}$

Step 7

Find the period of $\tan (\frac{θ}{2})$ .

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Replace $b$ with $\frac{1}{2}$ in the formula for period.

$\frac{π}{| \frac{1}{2} |}$

$\frac{1}{2}$ is approximately $0.5$ which is positive so remove the absolute value

$\frac{π}{\frac{1}{2}}$

Multiply the numerator by the reciprocal of the denominator.

$π \cdot 2$

Move $2$ to the left of $π$ .

$2 π$

Step 8

The period of the $\tan (\frac{θ}{2})$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = \frac{2 π}{3} + 2 π n, \frac{8 π}{3} + 2 π n$ , for any integer $n$

Step 9

Consolidate the answers.

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

Solve for θ tan(theta/2) = square root of 3

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