Trigonometry Examples

$\tan (x) + \frac{\sqrt{3}}{\tan (x)} < 1 + \sqrt{3}$

Step 1

Substitute $u$ for $\tan (x)$ .

$(u) + \frac{\sqrt{3}}{u} < 1 + \sqrt{3}$

Step 2

Find the LCD of the terms in the equation.

Tap for more steps...

Step 2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, u, 1, 1$

Step 2.2

The LCM of one and any expression is the expression.

$u$

Step 3

Multiply each term in $u + \frac{\sqrt{3}}{u} < 1 + \sqrt{3}$ by $u$ to eliminate the fractions.

Tap for more steps...

Step 3.1

Multiply each term in $u + \frac{\sqrt{3}}{u} < 1 + \sqrt{3}$ by $u$ .

$u \cdot u + \frac{\sqrt{3}}{u} u < 1 u + \sqrt{3} u$

Step 3.2

Simplify the left side.

Tap for more steps...

Step 3.2.1

Simplify each term.

Tap for more steps...

Step 3.2.1.1

Multiply $u$ by $u$ .

$u^{2} + \frac{\sqrt{3}}{u} u < 1 u + \sqrt{3} u$

Step 3.2.1.2

Cancel the common factor of $u$ .

Tap for more steps...

Step 3.2.1.2.1

Cancel the common factor.

$u^{2} + \frac{\sqrt{3}}{u} u < 1 u + \sqrt{3} u$

Step 3.2.1.2.2

Rewrite the expression.

$u^{2} + \sqrt{3} < 1 u + \sqrt{3} u$

Step 3.3

Simplify the right side.

Tap for more steps...

Step 3.3.1

Multiply $u$ by $1$ .

$u^{2} + \sqrt{3} < u + \sqrt{3} u$

Step 4

Solve the inequality.

Tap for more steps...

Step 4.1

Rewrite so $u$ is on the left side of the inequality.

$u + \sqrt{3} u > u^{2} + \sqrt{3}$

Step 4.2

Subtract $u^{2}$ from both sides of the inequality.

$u + \sqrt{3} u - u^{2} > \sqrt{3}$

Step 4.3

Convert the inequality to an equation.

$u + \sqrt{3} u - u^{2} = \sqrt{3}$

Step 4.4

Subtract $\sqrt{3}$ from both sides of the equation.

$u + \sqrt{3} u - u^{2} - \sqrt{3} = 0$

Step 4.5

Factor the left side of the equation.

Tap for more steps...

Step 4.5.1

Reorder terms.

$- u^{2} + u + u \sqrt{3} - \sqrt{3} = 0$

Step 4.5.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 4.5.2.1

Group the first two terms and the last two terms.

$- u^{2} + u + u \sqrt{3} - \sqrt{3} = 0$

Step 4.5.2.2

Factor out the greatest common factor (GCF) from each group.

$u (- u + 1) - \sqrt{3} (- u + 1) = 0$

Step 4.5.3

Factor the polynomial by factoring out the greatest common factor, $- u + 1$ .

$(- u + 1) (u - \sqrt{3}) = 0$

Step 4.6

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$- u + 1 = 0$

$u - \sqrt{3} = 0$

Step 4.7

Set $- u + 1$ equal to $0$ and solve for $u$ .

Tap for more steps...

Step 4.7.1

Set $- u + 1$ equal to $0$ .

$- u + 1 = 0$

Step 4.7.2

Solve $- u + 1 = 0$ for $u$ .

Tap for more steps...

Step 4.7.2.1

Subtract $1$ from both sides of the equation.

$- u = - 1$

Step 4.7.2.2

Divide each term in $- u = - 1$ by $- 1$ and simplify.

Tap for more steps...

Step 4.7.2.2.1

Divide each term in $- u = - 1$ by $- 1$ .

$\frac{- u}{- 1} = \frac{- 1}{- 1}$

Step 4.7.2.2.2

Simplify the left side.

Tap for more steps...

Step 4.7.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{u}{1} = \frac{- 1}{- 1}$

Step 4.7.2.2.2.2

Divide $u$ by $1$ .

$u = \frac{- 1}{- 1}$

Step 4.7.2.2.3

Simplify the right side.

Tap for more steps...

Step 4.7.2.2.3.1

Divide $- 1$ by $- 1$ .

$u = 1$

Step 4.8

Set $u - \sqrt{3}$ equal to $0$ and solve for $u$ .

Tap for more steps...

Step 4.8.1

Set $u - \sqrt{3}$ equal to $0$ .

$u - \sqrt{3} = 0$

Step 4.8.2

Add $\sqrt{3}$ to both sides of the equation.

$u = \sqrt{3}$

Step 4.9

The final solution is all the values that make $(- u + 1) (u - \sqrt{3}) = 0$ true.

$u = 1, \sqrt{3}$

Step 5

Substitute $\tan (x)$ for $u$ .

$\tan (x) = 1, \sqrt{3}$

Step 6

Set up each of the solutions to solve for $x$ .

$\tan (x) = 1$

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

Step 7

Solve for $x$ in $\tan (x) = 1$ .

Tap for more steps...

Step 7.1

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

$x = \arctan (1)$

Step 7.2

Simplify the right side.

Tap for more steps...

Step 7.2.1

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

$x = \frac{π}{4}$

Step 7.3

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 7.4

Simplify $π + \frac{π}{4}$ .

Tap for more steps...

Step 7.4.1

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

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

Step 7.4.2

Combine fractions.

Tap for more steps...

Step 7.4.2.1

Combine $π$ and $\frac{4}{4}$ .

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

Step 7.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Step 7.4.3

Simplify the numerator.

Tap for more steps...

Step 7.4.3.1

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Step 7.4.3.2

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

Step 7.5

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

Tap for more steps...

Step 7.5.1

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

$\frac{π}{| b |}$

Step 7.5.2

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

$\frac{π}{| 1 |}$

Step 7.5.3

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

$\frac{π}{1}$

Step 7.5.4

Divide $π$ by $1$ .

$π$

Step 7.6

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 8

Solve for $x$ in $\tan (x) = \sqrt{3}$ .

Tap for more steps...

Step 8.1

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

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

Step 8.2

Simplify the right side.

Tap for more steps...

Step 8.2.1

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

$x = \frac{π}{3}$

Step 8.3

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 8.4

Simplify $π + \frac{π}{3}$ .

Tap for more steps...

Step 8.4.1

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 8.4.2

Combine fractions.

Tap for more steps...

Step 8.4.2.1

Combine $π$ and $\frac{3}{3}$ .

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

Step 8.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 3 + π}{3}$

Step 8.4.3

Simplify the numerator.

Tap for more steps...

Step 8.4.3.1

Move $3$ to the left of $π$ .

$x = \frac{3 \cdot π + π}{3}$

Step 8.4.3.2

Add $3 π$ and $π$ .

$x = \frac{4 π}{3}$

Step 8.5

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

Tap for more steps...

Step 8.5.1

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

$\frac{π}{| b |}$

Step 8.5.2

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

$\frac{π}{| 1 |}$

Step 8.5.3

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

$\frac{π}{1}$

Step 8.5.4

Divide $π$ by $1$ .

$π$

Step 8.6

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 9

List all of the solutions.

$x = \frac{π}{4} + π n, \frac{5 π}{4} + π n, \frac{π}{3} + π n, \frac{4 π}{3} + π n$ , for any integer $n$

Step 10

Consolidate the solutions.

Tap for more steps...

Step 10.1

Consolidate $\frac{π}{4} + π n$ and $\frac{5 π}{4} + π n$ to $\frac{π}{4} + π n$ .

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

Step 10.2

Consolidate $\frac{π}{3} + π n$ and $\frac{4 π}{3} + π n$ to $\frac{π}{3} + π n$ .

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

Step 11

Find the domain of $\tan (x) + \sqrt{3} \cot (x) - 1 - \sqrt{3}$ .

Tap for more steps...

Step 11.1

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 11.2

Set the argument in $\cot (x)$ equal to $π n$ to find where the expression is undefined.

$x = π n$ , for any integer $n$

Step 11.3

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

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

Step 12

Use each root to create test intervals.

$\frac{π}{4} < x < \frac{π}{3}$

$\frac{π}{3} < x < \frac{5 π}{4}$

$\frac{5 π}{4} < x < \frac{4 π}{3}$

Step 13

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

Tap for more steps...

Step 13.1

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

Tap for more steps...

Step 13.1.1

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

$x = 1$

Step 13.1.2

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

$\tan (1) + \frac{\sqrt{3}}{\tan (1)} < 1 + \sqrt{3}$

Step 13.1.3

The left side $2.66954475$ is less than the right side $2.7320508$ , which means that the given statement is always true.

True

Step 13.2

Test a value on the interval $\frac{π}{3} < x < \frac{5 π}{4}$ to see if it makes the inequality true.

Tap for more steps...

Step 13.2.1

Choose a value on the interval $\frac{π}{3} < x < \frac{5 π}{4}$ and see if this value makes the original inequality true.

$x = 2$

Step 13.2.2

Replace $x$ with $2$ in the original inequality.

$\tan (2) + \frac{\sqrt{3}}{\tan (2)} < 1 + \sqrt{3}$

Step 13.2.3

The left side $- 2.97772599$ is less than the right side $2.7320508$ , which means that the given statement is always true.

True

Step 13.3

Test a value on the interval $\frac{5 π}{4} < x < \frac{4 π}{3}$ to see if it makes the inequality true.

Tap for more steps...

Step 13.3.1

Choose a value on the interval $\frac{5 π}{4} < x < \frac{4 π}{3}$ and see if this value makes the original inequality true.

$x = 4$

Step 13.3.2

Replace $x$ with $4$ in the original inequality.

$\tan (4) + \frac{\sqrt{3}}{\tan (4)} < 1 + \sqrt{3}$

Step 13.3.3

The left side $2.65377824$ is less than the right side $2.7320508$ , which means that the given statement is always true.

True

Step 13.4

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

$\frac{π}{4} < x < \frac{π}{3}$ True

$\frac{π}{3} < x < \frac{5 π}{4}$ True

$\frac{5 π}{4} < x < \frac{4 π}{3}$ True

$\frac{π}{4} < x < \frac{π}{3}$ True

$\frac{π}{3} < x < \frac{5 π}{4}$ True

$\frac{5 π}{4} < x < \frac{4 π}{3}$ True

Step 14

The solution consists of all of the true intervals.

$\frac{π}{4} + π n < x < \frac{π}{3} + π n$ or $\frac{π}{3} + π n < x < \frac{5 π}{4} + π n$ or $\frac{5 π}{4} + π n < x < \frac{4 π}{3} + π n$ , for any integer $n$

Step 15

Combine the intervals.

$\frac{π}{3} + π n < x < \frac{5 π}{4} + π n or \frac{π}{4} + π n < x < \frac{π}{3} + π n or \frac{5 π}{4} + π n < x < \frac{4 π}{3} + π n$ , for any integer $n$

Step 16