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Popular Trigonometry >

-1<= tan(x/2-pi/3)<= sqrt(3)

Solution

- 1 \leq tan (\frac{x}{2} - \frac{π}{3}) \leq 3

Solution

\frac{π}{6} + 2 πn \leq x \leq \frac{4 π}{3} + 2 πn

Interval Notation

[\frac{π}{6} + 2 πn, \frac{4 π}{3} + 2 πn]

Decimal

0.52359 \dots + 2 πn \leq x \leq 4.18879 \dots + 2 πn

Solution steps

- 1 \leq tan (\frac{x}{2} - \frac{π}{3}) \leq 3

a \leq u \leq b

then

a \leq u and u \leq b

- 1 \leq tan (\frac{x}{2} - \frac{π}{3}) and tan (\frac{x}{2} - \frac{π}{3}) \leq 3

- 1 \leq tan (\frac{x}{2} - \frac{π}{3}) : \frac{π}{6} + 2 πn \leq x < \frac{5 π}{3} + 2 πn

- 1 \leq tan (\frac{x}{2} - \frac{π}{3})

Switch sides

tan (\frac{x}{2} - \frac{π}{3}) \geq - 1

tan (x) \geq a

then

arctan (a) + πn \leq x < \frac{π}{2} + πn

arctan (- 1) + πn \leq (\frac{x}{2} - \frac{π}{3}) < \frac{π}{2} + πn

a \leq u < b

then

a \leq u and u < b

arctan (- 1) + πn \leq \frac{x}{2} - \frac{π}{3} and \frac{x}{2} - \frac{π}{3} < \frac{π}{2} + πn

arctan (- 1) + πn \leq \frac{x}{2} - \frac{π}{3} : x \geq 2 πn + \frac{π}{6}

arctan (- 1) + πn \leq \frac{x}{2} - \frac{π}{3}

Switch sides

\frac{x}{2} - \frac{π}{3} \geq arctan (- 1) + πn

Simplify

arctan (- 1) + πn : - \frac{π}{4} + πn

arctan (- 1) + πn

arctan (- 1) = - \frac{π}{4}

arctan (- 1)

Use the following property:

arctan (- x) = - arctan (x)

arctan (- 1) = - arctan (1)

= - arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

arctan (1)

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

= - \frac{π}{4}

= - \frac{π}{4} + πn

\frac{x}{2} - \frac{π}{3} \geq - \frac{π}{4} + πn

Move

\frac{π}{3}

to the right side

\frac{x}{2} - \frac{π}{3} \geq - \frac{π}{4} + πn

Add

\frac{π}{3}

to both sides

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} \geq - \frac{π}{4} + πn + \frac{π}{3}

Simplify

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} \geq - \frac{π}{4} + πn + \frac{π}{3}

Simplify

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} : \frac{x}{2}

\frac{x}{2} - \frac{π}{3} + \frac{π}{3}

Add similar elements:

- \frac{π}{3} + \frac{π}{3} \geq 0

= \frac{x}{2}

Simplify

- \frac{π}{4} + πn + \frac{π}{3} : πn + \frac{π}{12}

- \frac{π}{4} + πn + \frac{π}{3}

Group like terms

= πn - \frac{π}{4} + \frac{π}{3}

Least Common Multiplier of

4, 3 : 12

4, 3

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Multiply each factor the greatest number of times it occurs in either

4

3

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

12

For

\frac{π}{4} :

multiply the denominator and numerator by

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

For

\frac{π}{3} :

multiply the denominator and numerator by

4

\frac{π}{3} = \frac{π 4}{3 \cdot 4} = \frac{π 4}{12}

= - \frac{π 3}{12} + \frac{π 4}{12}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- π 3 + π 4}{12}

Add similar elements:

- 3 π + 4 π = π

= πn + \frac{π}{12}

\frac{x}{2} \geq πn + \frac{π}{12}

\frac{x}{2} \geq πn + \frac{π}{12}

\frac{x}{2} \geq πn + \frac{π}{12}

Multiply both sides by

2

\frac{x}{2} \geq πn + \frac{π}{12}

Multiply both sides by

2

\frac{2 x}{2} \geq 2 πn + 2 \cdot \frac{π}{12}

Simplify

\frac{2 x}{2} \geq 2 πn + 2 \cdot \frac{π}{12}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 πn + 2 \cdot \frac{π}{12} : 2 πn + \frac{π}{6}

2 πn + 2 \cdot \frac{π}{12}

2 \cdot \frac{π}{12} = \frac{π}{6}

2 \cdot \frac{π}{12}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{12}

Cancel the common factor:

2

= \frac{π}{6}

= 2 πn + \frac{π}{6}

x \geq 2 πn + \frac{π}{6}

x \geq 2 πn + \frac{π}{6}

x \geq 2 πn + \frac{π}{6}

\frac{x}{2} - \frac{π}{3} < \frac{π}{2} + πn : x < 2 πn + \frac{5 π}{3}

\frac{x}{2} - \frac{π}{3} < \frac{π}{2} + πn

Move

\frac{π}{3}

to the right side

\frac{x}{2} - \frac{π}{3} < \frac{π}{2} + πn

Add

\frac{π}{3}

to both sides

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} < \frac{π}{2} + πn + \frac{π}{3}

Simplify

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} < \frac{π}{2} + πn + \frac{π}{3}

Simplify

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} : \frac{x}{2}

\frac{x}{2} - \frac{π}{3} + \frac{π}{3}

Add similar elements:

- \frac{π}{3} + \frac{π}{3} < 0

= \frac{x}{2}

Simplify

\frac{π}{2} + πn + \frac{π}{3} : πn + \frac{5 π}{6}

\frac{π}{2} + πn + \frac{π}{3}

Group like terms

= πn + \frac{π}{2} + \frac{π}{3}

Least Common Multiplier of

2, 3 : 6

2, 3

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Multiply each factor the greatest number of times it occurs in either

2

3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

6

For

\frac{π}{2} :

multiply the denominator and numerator by

3

\frac{π}{2} = \frac{π 3}{2 \cdot 3} = \frac{π 3}{6}

For

\frac{π}{3} :

multiply the denominator and numerator by

2

\frac{π}{3} = \frac{π 2}{3 \cdot 2} = \frac{π 2}{6}

= \frac{π 3}{6} + \frac{π 2}{6}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π 3 + π 2}{6}

Add similar elements:

3 π + 2 π = 5 π

= πn + \frac{5 π}{6}

\frac{x}{2} < πn + \frac{5 π}{6}

\frac{x}{2} < πn + \frac{5 π}{6}

\frac{x}{2} < πn + \frac{5 π}{6}

Multiply both sides by

2

\frac{x}{2} < πn + \frac{5 π}{6}

Multiply both sides by

2

\frac{2 x}{2} < 2 πn + 2 \cdot \frac{5 π}{6}

Simplify

\frac{2 x}{2} < 2 πn + 2 \cdot \frac{5 π}{6}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 πn + 2 \cdot \frac{5 π}{6} : 2 πn + \frac{5 π}{3}

2 πn + 2 \cdot \frac{5 π}{6}

2 \cdot \frac{5 π}{6} = \frac{5 π}{3}

2 \cdot \frac{5 π}{6}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{5 π 2}{6}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{6}

Cancel the common factor:

2

= \frac{5 π}{3}

= 2 πn + \frac{5 π}{3}

x < 2 πn + \frac{5 π}{3}

x < 2 πn + \frac{5 π}{3}

x < 2 πn + \frac{5 π}{3}

Combine the intervals

x \geq 2 πn + \frac{π}{6} and x < 2 πn + \frac{5 π}{3}

Merge Overlapping Intervals

\frac{π}{6} + 2 πn \leq x < \frac{5 π}{3} + 2 πn

tan (\frac{x}{2} - \frac{π}{3}) \leq 3 : - \frac{π}{3} + 2 πn < x \leq \frac{4 π}{3} + 2 πn

tan (\frac{x}{2} - \frac{π}{3}) \leq 3

tan (x) \leq a

then

- \frac{π}{2} + πn < x \leq arctan (a) + πn

- \frac{π}{2} + πn < (\frac{x}{2} - \frac{π}{3}) \leq arctan (3) + πn

a < u \leq b

then

a < u and u \leq b

- \frac{π}{2} + πn < \frac{x}{2} - \frac{π}{3} and \frac{x}{2} - \frac{π}{3} \leq arctan (3) + πn

- \frac{π}{2} + πn < \frac{x}{2} - \frac{π}{3} : x > 2 πn - \frac{π}{3}

- \frac{π}{2} + πn < \frac{x}{2} - \frac{π}{3}

Switch sides

\frac{x}{2} - \frac{π}{3} > - \frac{π}{2} + πn

Move

\frac{π}{3}

to the right side

\frac{x}{2} - \frac{π}{3} > - \frac{π}{2} + πn

Add

\frac{π}{3}

to both sides

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} > - \frac{π}{2} + πn + \frac{π}{3}

Simplify

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} > - \frac{π}{2} + πn + \frac{π}{3}

Simplify

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} : \frac{x}{2}

\frac{x}{2} - \frac{π}{3} + \frac{π}{3}

Add similar elements:

- \frac{π}{3} + \frac{π}{3} > 0

= \frac{x}{2}

Simplify

- \frac{π}{2} + πn + \frac{π}{3} : πn - \frac{π}{6}

- \frac{π}{2} + πn + \frac{π}{3}

Group like terms

= πn - \frac{π}{2} + \frac{π}{3}

Least Common Multiplier of

2, 3 : 6

2, 3

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Multiply each factor the greatest number of times it occurs in either

2

3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

6

For

\frac{π}{2} :

multiply the denominator and numerator by

3

\frac{π}{2} = \frac{π 3}{2 \cdot 3} = \frac{π 3}{6}

For

\frac{π}{3} :

multiply the denominator and numerator by

2

\frac{π}{3} = \frac{π 2}{3 \cdot 2} = \frac{π 2}{6}

= - \frac{π 3}{6} + \frac{π 2}{6}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- π 3 + π 2}{6}

Add similar elements:

- 3 π + 2 π = - π

= \frac{- π}{6}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= πn - \frac{π}{6}

\frac{x}{2} > πn - \frac{π}{6}

\frac{x}{2} > πn - \frac{π}{6}

\frac{x}{2} > πn - \frac{π}{6}

Multiply both sides by

2

\frac{x}{2} > πn - \frac{π}{6}

Multiply both sides by

2

\frac{2 x}{2} > 2 πn - 2 \cdot \frac{π}{6}

Simplify

\frac{2 x}{2} > 2 πn - 2 \cdot \frac{π}{6}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 πn - 2 \cdot \frac{π}{6} : 2 πn - \frac{π}{3}

2 πn - 2 \cdot \frac{π}{6}

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{6}

Cancel the common factor:

2

= \frac{π}{3}

= 2 πn - \frac{π}{3}

x > 2 πn - \frac{π}{3}

x > 2 πn - \frac{π}{3}

x > 2 πn - \frac{π}{3}

\frac{x}{2} - \frac{π}{3} \leq arctan (3) + πn : x \leq \frac{4 π}{3} + 2 πn

\frac{x}{2} - \frac{π}{3} \leq arctan (3) + πn

Simplify

arctan (3) + πn : \frac{π}{3} + πn

arctan (3) + πn

Use the following trivial identity:

arctan (3) = \frac{π}{3}

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{3} + πn

\frac{x}{2} - \frac{π}{3} \leq \frac{π}{3} + πn

Move

\frac{π}{3}

to the right side

\frac{x}{2} - \frac{π}{3} \leq \frac{π}{3} + πn

Add

\frac{π}{3}

to both sides

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} \leq \frac{π}{3} + πn + \frac{π}{3}

Simplify

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} \leq \frac{π}{3} + πn + \frac{π}{3}

Simplify

\frac{x}{2} - \frac{π}{3} + \frac{π}{3} : \frac{x}{2}

\frac{x}{2} - \frac{π}{3} + \frac{π}{3}

Add similar elements:

- \frac{π}{3} + \frac{π}{3} \leq 0

= \frac{x}{2}

Simplify

\frac{π}{3} + πn + \frac{π}{3} : \frac{2 π}{3} + πn

\frac{π}{3} + πn + \frac{π}{3}

Group like terms

= \frac{π}{3} + \frac{π}{3} + πn

Combine the fractions

\frac{π}{3} + \frac{π}{3} : \frac{2 π}{3}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π + π}{3}

Add similar elements:

π + π = 2 π

= \frac{2 π}{3}

= \frac{2 π}{3} + πn

\frac{x}{2} \leq \frac{2 π}{3} + πn

\frac{x}{2} \leq \frac{2 π}{3} + πn

\frac{x}{2} \leq \frac{2 π}{3} + πn

Multiply both sides by

2

\frac{x}{2} \leq \frac{2 π}{3} + πn

Multiply both sides by

2

\frac{2 x}{2} \leq 2 \cdot \frac{2 π}{3} + 2 πn

Simplify

\frac{2 x}{2} \leq 2 \cdot \frac{2 π}{3} + 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 \cdot \frac{2 π}{3} + 2 πn : \frac{4 π}{3} + 2 πn

2 \cdot \frac{2 π}{3} + 2 πn

2 \cdot \frac{2 π}{3} = \frac{4 π}{3}

2 \cdot \frac{2 π}{3}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 π 2}{3}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 π}{3}

= \frac{4 π}{3} + 2 πn

x \leq \frac{4 π}{3} + 2 πn

x \leq \frac{4 π}{3} + 2 πn

x \leq \frac{4 π}{3} + 2 πn

Combine the intervals

x > 2 πn - \frac{π}{3} and x \leq \frac{4 π}{3} + 2 πn

Merge Overlapping Intervals

- \frac{π}{3} + 2 πn < x \leq \frac{4 π}{3} + 2 πn

Combine the intervals

\frac{π}{6} + 2 πn \leq x < \frac{5 π}{3} + 2 πn and - \frac{π}{3} + 2 πn < x \leq \frac{4 π}{3} + 2 πn

Merge Overlapping Intervals

\frac{π}{6} + 2 πn \leq x \leq \frac{4 π}{3} + 2 πn

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