Precalculus Examples

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

Step 1

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

$t = \arctan (\sqrt{3})$

Step 2

Simplify the right side.

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Step 2.1

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

$t = \frac{π}{3}$

Step 3

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.

$t = π + \frac{π}{3}$

Step 4

Simplify $π + \frac{π}{3}$ .

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Step 4.1

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

$t = π \cdot \frac{3}{3} + \frac{π}{3}$

Step 4.2

Combine fractions.

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Step 4.2.1

Combine $π$ and $\frac{3}{3}$ .

$t = \frac{π \cdot 3}{3} + \frac{π}{3}$

Step 4.2.2

Combine the numerators over the common denominator.

$t = \frac{π \cdot 3 + π}{3}$

Step 4.3

Simplify the numerator.

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Step 4.3.1

Move $3$ to the left of $π$ .

$t = \frac{3 \cdot π + π}{3}$

Step 4.3.2

Add $3 π$ and $π$ .

$t = \frac{4 π}{3}$

Step 5

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

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Step 5.1

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

$\frac{π}{| b |}$

Step 5.2

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

$\frac{π}{| 1 |}$

Step 5.3

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

$\frac{π}{1}$

Step 5.4

Divide $π$ by $1$ .

$π$

Step 6

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

$t = \frac{π}{3} + π n, \frac{4 π}{3} + π n$ , for any integer $n$

Step 7

Consolidate the answers.

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