Trigonometry Examples

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

Step 1

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

$x \geq \arctan (\sqrt{3})$

Step 2

Simplify the right side.

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Step 2.1

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

$x \geq \frac{π}{3}$

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.

$x = π + \frac{π}{3}$

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 $\frac{3}{3}$ .

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

Step 4.2

Combine fractions.

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Step 4.2.1

Combine $π$ and $\frac{3}{3}$ .

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

Step 4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 3 + π}{3}$

Step 4.3

Simplify the numerator.

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Step 4.3.1

Move $3$ to the left of $π$ .

$x = \frac{3 \cdot π + π}{3}$

Step 4.3.2

Add $3 π$ and $π$ .

$x = \frac{4 π}{3}$

Step 5

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

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

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

Step 7

Consolidate the answers.

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

Step 8

Find the domain of $\tan (x) - \sqrt{3}$ .

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Step 8.1

Set the argument in $\tan (x)$ equal to $\frac{π}{2} + π n$ to find where the expression is undefined.

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

Step 8.2

The domain is all values of $x$ that make the expression defined.

${x | x \neq \frac{π}{2} + π n}$ $n$ , for any integer $n$

Step 9

Use each root to create test intervals.

$\frac{π}{3} < x < \frac{π}{2}$

$\frac{π}{2} < x < \frac{4 π}{3}$

Step 10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 10.1

Test a value on the interval $\frac{π}{3} < x < \frac{π}{2}$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $\frac{π}{3} < x < \frac{π}{2}$ and see if this value makes the original inequality true.

$x = 1.31$

Step 10.1.2

Replace $x$ with $1.31$ in the original inequality.

$\tan (1.31) \geq \sqrt{3}$

Step 10.1.3

The left side $3.74708097$ is greater than the right side $1.7320508$ , which means that the given statement is always true.

True

Step 10.2

Test a value on the interval $\frac{π}{2} < x < \frac{4 π}{3}$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $\frac{π}{2} < x < \frac{4 π}{3}$ and see if this value makes the original inequality true.

$x = 3$

Step 10.2.2

Replace $x$ with $3$ in the original inequality.

$\tan (3) \geq \sqrt{3}$

Step 10.2.3

The left side $- 0.14254654$ is less than the right side $1.7320508$ , which means that the given statement is false.

False

Step 10.3

Compare the intervals to determine which ones satisfy the original inequality.

$\frac{π}{3} < x < \frac{π}{2}$ True

$\frac{π}{2} < x < \frac{4 π}{3}$ False

$\frac{π}{3} < x < \frac{π}{2}$ True

$\frac{π}{2} < x < \frac{4 π}{3}$ False

Step 11

The solution consists of all of the true intervals.

$\frac{π}{3} + π n \leq x < \frac{π}{2} + π n$ , for any integer $n$

Step 12

Convert the inequality to interval notation.

$[\frac{π}{3} + π n, \frac{π}{2} + π n)$

Step 13