Solve for x 3tan(x) = square root of 3

$3 \tan (x) = \sqrt{3}$

Step 1

Divide each term in $3 \tan (x) = \sqrt{3}$ by $3$ and simplify.

Divide each term in $3 \tan (x) = \sqrt{3}$ by $3$ .

$\frac{3 \tan (x)}{3} = \frac{\sqrt{3}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 \tan (x)}{3} = \frac{\sqrt{3}}{3}$

Divide $\tan (x)$ by $1$ .

$\tan (x) = \frac{\sqrt{3}}{3}$

Step 2

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

$x = \arctan (\frac{\sqrt{3}}{3})$

Step 3

Simplify the right side.

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

$x = \frac{π}{6}$

Step 4

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.

$x = π + \frac{π}{6}$

Step 5

Simplify $π + \frac{π}{6}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} + \frac{π}{6}$

Combine fractions.

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} + \frac{π}{6}$

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 + π}{6}$

Simplify the numerator.

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π + π}{6}$

Add $6 π$ and $π$ .

$x = \frac{7 π}{6}$

Step 6

Find the period of $\tan (x)$ .

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 $1$ .

$π$

Step 7

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

$x = \frac{π}{6} + π n, \frac{7 π}{6} + π n$ , for any integer $n$

Step 8

Consolidate the answers.

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

Solve for x 3tan(x) = square root of 3

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