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Popular Trigonometry >

tan(3x-1)>-sqrt(3)

Solution

tan (3 x - 1) > - 3

Solution

\frac{3 - π}{9} + \frac{π}{3} n < x < \frac{2 + π}{6} + \frac{π}{3} n

Interval Notation

(\frac{3 - π}{9} + \frac{π}{3} n, \frac{2 + π}{6} + \frac{π}{3} n)

Decimal

- 0.01573 \dots + \frac{π}{3} n < x < 0.85693 \dots + \frac{π}{3} n

Solution steps

tan (3 x - 1) > - 3

tan (x) > a

then

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

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

a < u < b

then

a < u and u < b

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

arctan (- 3) + πn < 3 x - 1 : x > \frac{3 - π}{9} + \frac{π}{3} n

arctan (- 3) + πn < 3 x - 1

Switch sides

3 x - 1 > arctan (- 3) + πn

Simplify

arctan (- 3) + πn : - \frac{π}{3} + πn

arctan (- 3) + πn

arctan (- 3) = - \frac{π}{3}

arctan (- 3)

Use the following property:

arctan (- x) = - arctan (x)

arctan (- 3) = - arctan (3)

= - arctan (3)

Use the following trivial identity:

arctan (3) = \frac{π}{3}

arctan (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}

= - \frac{π}{3}

= - \frac{π}{3} + πn

3 x - 1 > - \frac{π}{3} + πn

Move

1

to the right side

3 x - 1 > - \frac{π}{3} + πn

Add

1

to both sides

3 x - 1 + 1 > - \frac{π}{3} + πn + 1

Simplify

3 x > - \frac{π}{3} + πn + 1

3 x > - \frac{π}{3} + πn + 1

Divide both sides by

3

3 x > - \frac{π}{3} + πn + 1

Divide both sides by

3

\frac{3 x}{3} > - \frac{\frac{π}{3}}{3} + \frac{πn}{3} + \frac{1}{3}

Simplify

\frac{3 x}{3} > - \frac{\frac{π}{3}}{3} + \frac{πn}{3} + \frac{1}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

- \frac{\frac{π}{3}}{3} + \frac{πn}{3} + \frac{1}{3} : \frac{1}{3} - \frac{π}{9} + \frac{πn}{3}

- \frac{\frac{π}{3}}{3} + \frac{πn}{3} + \frac{1}{3}

Group like terms

= \frac{1}{3} + \frac{πn}{3} - \frac{\frac{π}{3}}{3}

\frac{\frac{π}{3}}{3} = \frac{π}{9}

\frac{\frac{π}{3}}{3}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{π}{9}

= \frac{1}{3} + \frac{πn}{3} - \frac{π}{9}

Group like terms

= \frac{1}{3} - \frac{π}{9} + \frac{πn}{3}

x > \frac{1}{3} - \frac{π}{9} + \frac{πn}{3}

x > \frac{1}{3} - \frac{π}{9} + \frac{πn}{3}

Simplify

\frac{1}{3} - \frac{π}{9} : \frac{3 - π}{9}

\frac{1}{3} - \frac{π}{9}

Least Common Multiplier of

3, 9 : 9

3, 9

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

9 : 3 \cdot 3

9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3

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

3

9

= 3 \cdot 3

Multiply the numbers:

3 \cdot 3 = 9

= 9

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

9

For

\frac{1}{3} :

multiply the denominator and numerator by

3

\frac{1}{3} = \frac{1 \cdot 3}{3 \cdot 3} = \frac{3}{9}

= \frac{3}{9} - \frac{π}{9}

Since the denominators are equal, combine the fractions:

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

= \frac{3 - π}{9}

x > \frac{3 - π}{9} + \frac{π}{3} n

x > \frac{3 - π}{9} + \frac{π}{3} n

3 x - 1 < \frac{π}{2} + πn : x < \frac{2 + π}{6} + \frac{π}{3} n

3 x - 1 < \frac{π}{2} + πn

Move

1

to the right side

3 x - 1 < \frac{π}{2} + πn

Add

1

to both sides

3 x - 1 + 1 < \frac{π}{2} + πn + 1

Simplify

3 x < \frac{π}{2} + πn + 1

3 x < \frac{π}{2} + πn + 1

Divide both sides by

3

3 x < \frac{π}{2} + πn + 1

Divide both sides by

3

\frac{3 x}{3} < \frac{\frac{π}{2}}{3} + \frac{πn}{3} + \frac{1}{3}

Simplify

\frac{3 x}{3} < \frac{\frac{π}{2}}{3} + \frac{πn}{3} + \frac{1}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{π}{2}}{3} + \frac{πn}{3} + \frac{1}{3} : \frac{1}{3} + \frac{π}{6} + \frac{πn}{3}

\frac{\frac{π}{2}}{3} + \frac{πn}{3} + \frac{1}{3}

Group like terms

= \frac{1}{3} + \frac{πn}{3} + \frac{\frac{π}{2}}{3}

\frac{\frac{π}{2}}{3} = \frac{π}{6}

\frac{\frac{π}{2}}{3}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{π}{6}

= \frac{1}{3} + \frac{πn}{3} + \frac{π}{6}

Group like terms

= \frac{1}{3} + \frac{π}{6} + \frac{πn}{3}

x < \frac{1}{3} + \frac{π}{6} + \frac{πn}{3}

x < \frac{1}{3} + \frac{π}{6} + \frac{πn}{3}

Simplify

\frac{1}{3} + \frac{π}{6} : \frac{2 + π}{6}

\frac{1}{3} + \frac{π}{6}

Least Common Multiplier of

3, 6 : 6

3, 6

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

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

3

6

= 3 \cdot 2

Multiply the numbers:

3 \cdot 2 = 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{1}{3} :

multiply the denominator and numerator by

2

\frac{1}{3} = \frac{1 \cdot 2}{3 \cdot 2} = \frac{2}{6}

= \frac{2}{6} + \frac{π}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{2 + π}{6}

x < \frac{2 + π}{6} + \frac{π}{3} n

x < \frac{2 + π}{6} + \frac{π}{3} n

Combine the intervals

x > \frac{3 - π}{9} + \frac{π}{3} n and x < \frac{2 + π}{6} + \frac{π}{3} n

Merge Overlapping Intervals

\frac{3 - π}{9} + \frac{π}{3} n < x < \frac{2 + π}{6} + \frac{π}{3} n

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