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Popular >

tan(x-pi/5)>-sqrt(3)

Solution

tan (x - \frac{π}{5}) > - 3

Solution

- \frac{2 π}{15} + πn < x < \frac{7 π}{10} + πn

Interval Notation

(- \frac{2 π}{15} + πn, \frac{7 π}{10} + πn)

Decimal Notation

- 0.41887 \dots + πn < x < 2.19911 \dots + πn

Solution steps

tan (x - \frac{π}{5}) > - 3

tan (x) > a

then

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

arctan (- 3) + πn < (x - \frac{π}{5}) < \frac{π}{2} + πn

a < u < b

then

a < u and u < b

arctan (- 3) + πn < x - \frac{π}{5} and x - \frac{π}{5} < \frac{π}{2} + πn

arctan (- 3) + πn < x - \frac{π}{5} : x > πn - \frac{2 π}{15}

arctan (- 3) + πn < x - \frac{π}{5}

Switch sides

x - \frac{π}{5} > 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

x - \frac{π}{5} > - \frac{π}{3} + πn

Move

\frac{π}{5}

to the right side

x - \frac{π}{5} > - \frac{π}{3} + πn

Add

\frac{π}{5}

to both sides

x - \frac{π}{5} + \frac{π}{5} > - \frac{π}{3} + πn + \frac{π}{5}

Simplify

x - \frac{π}{5} + \frac{π}{5} > - \frac{π}{3} + πn + \frac{π}{5}

Simplify

x - \frac{π}{5} + \frac{π}{5} : x

x - \frac{π}{5} + \frac{π}{5}

Add similar elements:

- \frac{π}{5} + \frac{π}{5} > 0

= x

Simplify

- \frac{π}{3} + πn + \frac{π}{5} : πn - \frac{2 π}{15}

- \frac{π}{3} + πn + \frac{π}{5}

Group like terms

= πn - \frac{π}{3} + \frac{π}{5}

Least Common Multiplier of

3, 5 : 15

3, 5

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

5 : 5

5

5

is a prime number, therefore no factorization is possible

= 5

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

3

5

= 3 \cdot 5

Multiply the numbers:

3 \cdot 5 = 15

= 15

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

15

For

\frac{π}{3} :

multiply the denominator and numerator by

5

\frac{π}{3} = \frac{π 5}{3 \cdot 5} = \frac{π 5}{15}

For

\frac{π}{5} :

multiply the denominator and numerator by

3

\frac{π}{5} = \frac{π 3}{5 \cdot 3} = \frac{π 3}{15}

= - \frac{π 5}{15} + \frac{π 3}{15}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 5 + π 3}{15}

Add similar elements:

- 5 π + 3 π = - 2 π

= \frac{- 2 π}{15}

Apply the fraction rule:

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

= πn - \frac{2 π}{15}

x > πn - \frac{2 π}{15}

x > πn - \frac{2 π}{15}

x > πn - \frac{2 π}{15}

x - \frac{π}{5} < \frac{π}{2} + πn : x < πn + \frac{7 π}{10}

x - \frac{π}{5} < \frac{π}{2} + πn

Move

\frac{π}{5}

to the right side

x - \frac{π}{5} < \frac{π}{2} + πn

Add

\frac{π}{5}

to both sides

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

Simplify

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

Simplify

x - \frac{π}{5} + \frac{π}{5} : x

x - \frac{π}{5} + \frac{π}{5}

Add similar elements:

- \frac{π}{5} + \frac{π}{5} < 0

= x

Simplify

\frac{π}{2} + πn + \frac{π}{5} : πn + \frac{7 π}{10}

\frac{π}{2} + πn + \frac{π}{5}

Group like terms

= πn + \frac{π}{2} + \frac{π}{5}

Least Common Multiplier of

2, 5 : 10

2, 5

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

5 : 5

5

5

is a prime number, therefore no factorization is possible

= 5

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

2

5

= 2 \cdot 5

Multiply the numbers:

2 \cdot 5 = 10

= 10

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

10

For

\frac{π}{2} :

multiply the denominator and numerator by

5

\frac{π}{2} = \frac{π 5}{2 \cdot 5} = \frac{π 5}{10}

For

\frac{π}{5} :

multiply the denominator and numerator by

2

\frac{π}{5} = \frac{π 2}{5 \cdot 2} = \frac{π 2}{10}

= \frac{π 5}{10} + \frac{π 2}{10}

Since the denominators are equal, combine the fractions:

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

= \frac{π 5 + π 2}{10}

Add similar elements:

5 π + 2 π = 7 π

= πn + \frac{7 π}{10}

x < πn + \frac{7 π}{10}

x < πn + \frac{7 π}{10}

x < πn + \frac{7 π}{10}

Combine the intervals

x > πn - \frac{2 π}{15} and x < πn + \frac{7 π}{10}

Merge Overlapping Intervals

- \frac{2 π}{15} + πn < x < \frac{7 π}{10} + πn

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