Trigonometry Examples

$\tan (x) < 2 \sin (x)$

Step 1

Divide each term in $\tan (x) < 2 \sin (x)$ by $\tan (x)$ and simplify.

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Step 1.1

Divide each term in $\tan (x) < 2 \sin (x)$ by $\tan (x)$ .

$\frac{\tan (x)}{\tan (x)} < \frac{2 \sin (x)}{\tan (x)}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $\tan (x)$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{\tan (x)}{\tan (x)} < \frac{2 \sin (x)}{\tan (x)}$

Step 1.2.1.2

Rewrite the expression.

$1 < \frac{2 \sin (x)}{\tan (x)}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Separate fractions.

$1 < \frac{2}{1} \cdot \frac{\sin (x)}{\tan (x)}$

Step 1.3.2

Rewrite $\tan (x)$ in terms of sines and cosines.

$1 < \frac{2}{1} \cdot \frac{\sin (x)}{\frac{\sin (x)}{\cos (x)}}$

Step 1.3.3

Multiply by the reciprocal of the fraction to divide by $\frac{\sin (x)}{\cos (x)}$ .

$1 < \frac{2}{1} (\sin (x) \frac{\cos (x)}{\sin (x)})$

Step 1.3.4

Write $\sin (x)$ as a fraction with denominator $1$ .

$1 < \frac{2}{1} (\frac{\sin (x)}{1} \cdot \frac{\cos (x)}{\sin (x)})$

Step 1.3.5

Cancel the common factor of $\sin (x)$ .

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Step 1.3.5.1

Cancel the common factor.

$1 < \frac{2}{1} (\frac{\sin (x)}{1} \cdot \frac{\cos (x)}{\sin (x)})$

Step 1.3.5.2

Rewrite the expression.

$1 < \frac{2}{1} \cos (x)$

Step 1.3.6

Divide $2$ by $1$ .

$1 < 2 \cos (x)$

Step 2

Rewrite so $\cos (x)$ is on the left side of the inequality.

$2 \cos (x) > 1$

Step 3

Divide each term in $2 \cos (x) > 1$ by $2$ and simplify.

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Step 3.1

Divide each term in $2 \cos (x) > 1$ by $2$ .

$\frac{2 \cos (x)}{2} > \frac{1}{2}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $2$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{2 \cos (x)}{2} > \frac{1}{2}$

Step 3.2.1.2

Divide $\cos (x)$ by $1$ .

$\cos (x) > \frac{1}{2}$

Step 4

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

$x > \arccos (\frac{1}{2})$

Step 5

Simplify the right side.

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Step 5.1

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$x > \frac{π}{3}$

Step 6

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - \frac{π}{3}$

Step 7

Simplify $2 π - \frac{π}{3}$ .

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Step 7.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = 2 π \cdot \frac{3}{3} - \frac{π}{3}$

Step 7.2

Combine fractions.

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Step 7.2.1

Combine $2 π$ and $\frac{3}{3}$ .

$x = \frac{2 π \cdot 3}{3} - \frac{π}{3}$

Step 7.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 7.3

Simplify the numerator.

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Step 7.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 7.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 8

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

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Step 8.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 8.2

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

$\frac{2 π}{| 1 |}$

Step 8.3

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

$\frac{2 π}{1}$

Step 8.4

Divide $2 π$ by $1$ .

$2 π$

Step 9

The period of the $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 10

Find the domain of $\tan (x) - 2 \sin (x)$ .

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Step 10.1

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 10.2

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 11

Use each root to create test intervals.

$\frac{π}{3} < x < \frac{π}{2}$

$\frac{π}{2} < x < \frac{3 π}{2}$

$\frac{3 π}{2} < x < \frac{5 π}{3}$

$\frac{5 π}{3} < x < \frac{7 π}{3}$

$\frac{7 π}{3} < x < \frac{5 π}{2}$

$\frac{5 π}{2} < x < \frac{7 π}{2}$

$\frac{7 π}{2} < x < \frac{11 π}{3}$

Step 12

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 12.1

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 12.1.1

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 12.1.2

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

$\tan (1.31) < 2 \sin (1.31)$

Step 12.1.3

The left side $3.74708097$ is not less than the right side $1.9323699$ , which means that the given statement is false.

False

Step 12.2

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

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Step 12.2.1

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

$x = 3$

Step 12.2.2

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

$\tan (3) < 2 \sin (3)$

Step 12.2.3

The left side $- 0.14254654$ is less than the right side $0.28224001$ , which means that the given statement is always true.

True

Step 12.3

Test a value on the interval $\frac{3 π}{2} < x < \frac{5 π}{3}$ to see if it makes the inequality true.

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Step 12.3.1

Choose a value on the interval $\frac{3 π}{2} < x < \frac{5 π}{3}$ and see if this value makes the original inequality true.

$x = 5$

Step 12.3.2

Replace $x$ with $5$ in the original inequality.

$\tan (5) < 2 \sin (5)$

Step 12.3.3

The left side $- 3.380515$ is less than the right side $- 1.91784854$ , which means that the given statement is always true.

True

Step 12.4

Test a value on the interval $\frac{5 π}{3} < x < \frac{7 π}{3}$ to see if it makes the inequality true.

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Step 12.4.1

Choose a value on the interval $\frac{5 π}{3} < x < \frac{7 π}{3}$ and see if this value makes the original inequality true.

$x = 6$

Step 12.4.2

Replace $x$ with $6$ in the original inequality.

$\tan (6) < 2 \sin (6)$

Step 12.4.3

The left side $- 0.29100619$ is not less than the right side $- 0.55883099$ , which means that the given statement is false.

False

Step 12.5

Test a value on the interval $\frac{7 π}{3} < x < \frac{5 π}{2}$ to see if it makes the inequality true.

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Step 12.5.1

Choose a value on the interval $\frac{7 π}{3} < x < \frac{5 π}{2}$ and see if this value makes the original inequality true.

$x = 7.59$

Step 12.5.2

Replace $x$ with $7.59$ in the original inequality.

$\tan (7.59) < 2 \sin (7.59)$

Step 12.5.3

The left side $3.69973691$ is not less than the right side $1.93071743$ , which means that the given statement is false.

False

Step 12.6

Test a value on the interval $\frac{5 π}{2} < x < \frac{7 π}{2}$ to see if it makes the inequality true.

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Step 12.6.1

Choose a value on the interval $\frac{5 π}{2} < x < \frac{7 π}{2}$ and see if this value makes the original inequality true.

$x = 9$

Step 12.6.2

Replace $x$ with $9$ in the original inequality.

$\tan (9) < 2 \sin (9)$

Step 12.6.3

The left side $- 0.45231565$ is less than the right side $0.82423697$ , which means that the given statement is always true.

True

Step 12.7

Test a value on the interval $\frac{7 π}{2} < x < \frac{11 π}{3}$ to see if it makes the inequality true.

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Step 12.7.1

Choose a value on the interval $\frac{7 π}{2} < x < \frac{11 π}{3}$ and see if this value makes the original inequality true.

$x = 11$

Step 12.7.2

Replace $x$ with $11$ in the original inequality.

$\tan (11) < 2 \sin (11)$

Step 12.7.3

The left side $- 225.95084645$ is less than the right side $- 1.99998041$ , which means that the given statement is always true.

True

Step 12.8

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

$\frac{π}{3} < x < \frac{π}{2}$ False

$\frac{π}{2} < x < \frac{3 π}{2}$ True

$\frac{3 π}{2} < x < \frac{5 π}{3}$ True

$\frac{5 π}{3} < x < \frac{7 π}{3}$ False

$\frac{7 π}{3} < x < \frac{5 π}{2}$ False

$\frac{5 π}{2} < x < \frac{7 π}{2}$ True

$\frac{7 π}{2} < x < \frac{11 π}{3}$ True

$\frac{π}{3} < x < \frac{π}{2}$ False

$\frac{π}{2} < x < \frac{3 π}{2}$ True

$\frac{3 π}{2} < x < \frac{5 π}{3}$ True

$\frac{5 π}{3} < x < \frac{7 π}{3}$ False

$\frac{7 π}{3} < x < \frac{5 π}{2}$ False

$\frac{5 π}{2} < x < \frac{7 π}{2}$ True

$\frac{7 π}{2} < x < \frac{11 π}{3}$ True

Step 13

The solution consists of all of the true intervals.

$\frac{π}{2} + 2 π n < x < \frac{3 π}{2} + 2 π n$ or $\frac{3 π}{2} + 2 π n < x < \frac{5 π}{3} + 2 π n$ or $\frac{5 π}{2} + 2 π n < x < \frac{7 π}{2} + 2 π n$ or $\frac{7 π}{2} + 2 π n < x < \frac{11 π}{3} + 2 π n$ , for any integer $n$

Step 14

Combine the intervals.

$\frac{5 π}{2} + 2 π n < x < \frac{7 π}{2} + 2 π n or \frac{π}{2} + 2 π n < x < \frac{3 π}{2} + 2 π n or \frac{3 π}{2} + 2 π n < x < \frac{5 π}{3} + 2 π n or \frac{7 π}{2} + 2 π n < x < \frac{11 π}{3} + 2 π n$ , for any integer $n$

Step 15