Trigonometry Examples

$\sqrt{3} \tan (θ) + 1 = 0$

Step 1

Subtract $1$ from both sides of the equation.

$\sqrt{3} \tan (θ) = - 1$

Step 2

Divide each term in $\sqrt{3} \tan (θ) = - 1$ by $\sqrt{3}$ and simplify.

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Step 2.1

Divide each term in $\sqrt{3} \tan (θ) = - 1$ by $\sqrt{3}$ .

$\frac{\sqrt{3} \tan (θ)}{\sqrt{3}} = \frac{- 1}{\sqrt{3}}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $\sqrt{3}$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{\sqrt{3} \tan (θ)}{\sqrt{3}} = \frac{- 1}{\sqrt{3}}$

Step 2.2.1.2

Divide $\tan (θ)$ by $1$ .

$\tan (θ) = \frac{- 1}{\sqrt{3}}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Move the negative in front of the fraction.

$\tan (θ) = - \frac{1}{\sqrt{3}}$

Step 2.3.2

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\tan (θ) = - (\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}})$

Step 2.3.3

Combine and simplify the denominator.

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Step 2.3.3.1

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

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

Step 2.3.3.2

Raise $\sqrt{3}$ to the power of $1$ .

$\tan (θ) = - \frac{\sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Step 2.3.3.3

Raise $\sqrt{3}$ to the power of $1$ .

$\tan (θ) = - \frac{\sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

Step 2.3.3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\tan (θ) = - \frac{\sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Step 2.3.3.5

Add $1$ and $1$ .

$\tan (θ) = - \frac{\sqrt{3}}{{\sqrt{3}}^{2}}$

Step 2.3.3.6

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 2.3.3.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$\tan (θ) = - \frac{\sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

Step 2.3.3.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\tan (θ) = - \frac{\sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Step 2.3.3.6.3

Combine $\frac{1}{2}$ and $2$ .

$\tan (θ) = - \frac{\sqrt{3}}{3^{\frac{2}{2}}}$

Step 2.3.3.6.4

Cancel the common factor of $2$ .

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Step 2.3.3.6.4.1

Cancel the common factor.

$\tan (θ) = - \frac{\sqrt{3}}{3^{\frac{2}{2}}}$

Step 2.3.3.6.4.2

Rewrite the expression.

$\tan (θ) = - \frac{\sqrt{3}}{3^{1}}$

Step 2.3.3.6.5

Evaluate the exponent.

$\tan (θ) = - \frac{\sqrt{3}}{3}$

Step 3

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

$θ = \arctan (- \frac{\sqrt{3}}{3})$

Step 4

Simplify the right side.

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Step 4.1

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

$θ = - \frac{π}{6}$

Step 5

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the third quadrant.

$θ = - \frac{π}{6} - π$

Step 6

Simplify the expression to find the second solution.

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Step 6.1

Add $2 π$ to $- \frac{π}{6} - π$ .

$θ = - \frac{π}{6} - π + 2 π$

Step 6.2

The resulting angle of $\frac{5 π}{6}$ is positive and coterminal with $- \frac{π}{6} - π$ .

$θ = \frac{5 π}{6}$

Step 7

Find the period of $\tan (θ)$ .

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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

Add $π$ to every negative angle to get positive angles.

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Step 8.1

Add $π$ to $- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + π$

Step 8.2

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

$π \cdot \frac{6}{6} - \frac{π}{6}$

Step 8.3

Combine fractions.

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Step 8.3.1

Combine $π$ and $\frac{6}{6}$ .

$\frac{π \cdot 6}{6} - \frac{π}{6}$

Step 8.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 6 - π}{6}$

Step 8.4

Simplify the numerator.

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Step 8.4.1

Move $6$ to the left of $π$ .

$\frac{6 \cdot π - π}{6}$

Step 8.4.2

Subtract $π$ from $6 π$ .

$\frac{5 π}{6}$

Step 8.5

List the new angles.

$θ = \frac{5 π}{6}$

Step 9

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

$θ = \frac{5 π}{6} + π n, \frac{5 π}{6} + π n$ , for any integer $n$

Step 10

Consolidate the answers.

$θ = \frac{5 π}{6} + π n$ , for any integer $n$