Precalculus Examples

$\tan (2 x) - \sqrt{3} = 0$

Step 1

Add $\sqrt{3}$ to both sides of the equation.

$\tan (2 x) = \sqrt{3}$

Step 2

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

$2 x = \arctan (\sqrt{3})$

Step 3

Simplify the right side.

Tap for more steps...

Step 3.1

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

$2 x = \frac{π}{3}$

Step 4

Divide each term in $2 x = \frac{π}{3}$ by $2$ and simplify.

Tap for more steps...

Step 4.1

Divide each term in $2 x = \frac{π}{3}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{π}{3}}{2}$

Step 4.2

Simplify the left side.

Tap for more steps...

Step 4.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 4.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{π}{3}}{2}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{3}}{2}$

Step 4.3

Simplify the right side.

Tap for more steps...

Step 4.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{3} \cdot \frac{1}{2}$

Step 4.3.2

Multiply $\frac{π}{3} \cdot \frac{1}{2}$ .

Tap for more steps...

Step 4.3.2.1

Multiply $\frac{π}{3}$ by $\frac{1}{2}$ .

$x = \frac{π}{3 \cdot 2}$

Step 4.3.2.2

Multiply $3$ by $2$ .

$x = \frac{π}{6}$

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.

$2 x = π + \frac{π}{3}$

Step 6

Solve for $x$ .

Tap for more steps...

Step 6.1

Simplify.

Tap for more steps...

Step 6.1.1

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

$2 x = π \cdot \frac{3}{3} + \frac{π}{3}$

Step 6.1.2

Combine $π$ and $\frac{3}{3}$ .

$2 x = \frac{π \cdot 3}{3} + \frac{π}{3}$

Step 6.1.3

Combine the numerators over the common denominator.

$2 x = \frac{π \cdot 3 + π}{3}$

Step 6.1.4

Add $π \cdot 3$ and $π$ .

Tap for more steps...

Step 6.1.4.1

Reorder $π$ and $3$ .

$2 x = \frac{3 \cdot π + π}{3}$

Step 6.1.4.2

Add $3 \cdot π$ and $π$ .

$2 x = \frac{4 \cdot π}{3}$

Step 6.2

Divide each term in $2 x = \frac{4 \cdot π}{3}$ by $2$ and simplify.

Tap for more steps...

Step 6.2.1

Divide each term in $2 x = \frac{4 \cdot π}{3}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{4 \cdot π}{3}}{2}$

Step 6.2.2

Simplify the left side.

Tap for more steps...

Step 6.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 6.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{4 \cdot π}{3}}{2}$

Step 6.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{4 \cdot π}{3}}{2}$

Step 6.2.3

Simplify the right side.

Tap for more steps...

Step 6.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{4 \cdot π}{3} \cdot \frac{1}{2}$

Step 6.2.3.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 6.2.3.2.1

Factor $2$ out of $4 \cdot π$ .

$x = \frac{2 (2 \cdot π)}{3} \cdot \frac{1}{2}$

Step 6.2.3.2.2

Cancel the common factor.

$x = \frac{2 (2 \cdot π)}{3} \cdot \frac{1}{2}$

Step 6.2.3.2.3

Rewrite the expression.

$x = \frac{2 \cdot π}{3}$

$x = \frac{2 π}{3}$

Step 7

Find the period of $\tan (2 x)$ .

Tap for more steps...

Step 7.1

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

$\frac{π}{| b |}$

Step 7.2

Replace $b$ with $2$ in the formula for period.

$\frac{π}{| 2 |}$

Step 7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{π}{2}$

Step 8

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

$x = \frac{π}{6} + \frac{π n}{2}, \frac{2 π}{3} + \frac{π n}{2}$ , for any integer $n$

Step 9

Consolidate the answers.

$x = \frac{π}{6} + \frac{π n}{2}$ , for any integer $n$