Solve the Inequality for x tan(x)<1

$\tan (x) < 1$

Step 1

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

$x < \arctan (1)$

Step 2

Simplify the right side.

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$x < \frac{π}{4}$

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.

$x = π + \frac{π}{4}$

Step 4

Simplify $π + \frac{π}{4}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = π \cdot \frac{4}{4} + \frac{π}{4}$

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

$x = \frac{π \cdot 4}{4} + \frac{π}{4}$

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

Step 5

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

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

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.

$x = \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$

Step 7

Consolidate the answers.

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

Step 8

Find the domain of $\tan (x) - 1$ .

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 $n$

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.

$\frac{π}{4} < x < \frac{π}{2}$

$\frac{π}{2} < x < \frac{5 π}{4}$

Step 10

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

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

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

$x = 1$

Replace $x$ with $1$ in the original inequality.

$\tan (1) < 1$

The left side $1.55740772$ is not less than the right side $1$ , which means that the given statement is false.

False

Test a value on the interval $\frac{π}{2} < x < \frac{5 π}{4}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{π}{2} < x < \frac{5 π}{4}$ and see if this value makes the original inequality true.

$x = 3$

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

$\tan (3) < 1$

The left side $- 0.14254654$ is less than the right side $1$ , which means that the given statement is always true.

True

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

$\frac{π}{4} < x < \frac{π}{2}$ False

$\frac{π}{2} < x < \frac{5 π}{4}$ True

$\frac{π}{4} < x < \frac{π}{2}$ False

$\frac{π}{2} < x < \frac{5 π}{4}$ True

Step 11

The solution consists of all of the true intervals.

$\frac{π}{2} + π n < x < \frac{5 π}{4} + π n$ , for any integer $n$

Step 12

Solve the Inequality for x tan(x)<1

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