Trigonometry Examples

$\tan (x) = 4 \tan (x) - \sqrt{3}$

Step 1

Move all terms containing $\tan (x)$ to the left side of the equation.

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Step 1.1

Subtract $4 \tan (x)$ from both sides of the equation.

$\tan (x) - 4 \tan (x) = - \sqrt{3}$

Step 1.2

Subtract $4 \tan (x)$ from $\tan (x)$ .

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $- 3$ .

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Step 2.2.1.1

Cancel the common factor.

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

Step 2.2.1.2

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

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

Step 2.3

Simplify the right side.

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Step 2.3.1

Dividing two negative values results in a positive value.

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

Step 3

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

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

Step 4

Simplify the right side.

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Step 4.1

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

$x = \frac{π}{6}$

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.

$x = π + \frac{π}{6}$

Step 6

Simplify $π + \frac{π}{6}$ .

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Step 6.1

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

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

Step 6.2

Combine fractions.

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Step 6.2.1

Combine $π$ and $\frac{6}{6}$ .

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

Step 6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 + π}{6}$

Step 6.3

Simplify the numerator.

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Step 6.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π + π}{6}$

Step 6.3.2

Add $6 π$ and $π$ .

$x = \frac{7 π}{6}$

Step 7

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

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Step 7.1

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

$\frac{π}{| b |}$

Step 7.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 7.4

Divide $π$ by $1$ .

$π$

Step 8

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 9

Consolidate the answers.

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