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Popular >

(3tan(x)-tan^3(x))/(1-tan^2(x))>0

Solution

\frac{3 tan ( x ) - tan ^{3} ( x )}{1 - tan ^{2} ( x )} > 0

Solution

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

Interval Notation

(πn, \frac{π}{4} + πn) \cup (\frac{π}{3} + πn, \frac{π}{2} + πn) \cup (\frac{2 π}{3} + πn, \frac{3 π}{4} + πn)

Decimal Notation

πn < x < 0.78539 \dots + πn or 1.04719 \dots + πn < x < 1.57079 \dots + πn or 2.09439 \dots + πn < x < 2.35619 \dots + πn

Solution steps

\frac{3 tan ( x ) - tan ^{3} ( x )}{1 - tan ^{2} ( x )} > 0

Let:

u = tan (x)

\frac{3 u - u ^{3}}{1 - u ^{2}} > 0

\frac{3 u - u ^{3}}{1 - u ^{2}} > 0 : - 3 < u < - 1 or 0 < u < 1 or u > 3

\frac{3 u - u ^{3}}{1 - u ^{2}} > 0

Factor

\frac{3 u - u ^{3}}{1 - u ^{2}} : \frac{- u ( u + 3 ) ( u - 3 )}{- ( u + 1 ) ( u - 1 )}

\frac{3 u - u ^{3}}{1 - u ^{2}}

Factor

- u^{2} + 1 : - (u + 1) (u - 1)

- u^{2} + 1

Factor out common term

- 1

= - (u^{2} - 1)

Factor

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

Rewrite

1

1^{2}

= u^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - 1^{2} = (u + 1) (u - 1)

= (u + 1) (u - 1)

= - (u + 1) (u - 1)

= \frac{3 u - u ^{3}}{- ( u + 1 ) ( u - 1 )}

Factor

- u^{3} + 3 u : - u (u + 3) (u - 3)

- u^{3} + 3 u

Factor out common term

- u : - u (u^{2} - 3)

- u^{3} + 3 u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{3} = u^{2} u

= - u^{2} u + 3 u

Factor out common term

- u

= - u (u^{2} - 3)

= - u (u^{2} - 3)

Factor

u^{2} - 3 : (u + 3) (u - 3)

u^{2} - 3

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

= u^{2} - (3)^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - (3)^{2} = (u + 3) (u - 3)

= (u + 3) (u - 3)

= - u (u + 3) (u - 3)

= \frac{- u ( u + 3 ) ( u - 3 )}{- ( u + 1 ) ( u - 1 )}

\frac{- u ( u + 3 ) ( u - 3 )}{- ( u + 1 ) ( u - 1 )} > 0

Identify the intervals

Find the signs of the factors of

\frac{- u ( u + 3 ) ( u - 3 )}{- ( u + 1 ) ( u - 1 )}

Find the signs of

u

u = 0

u < 0

u > 0

Find the signs of

u + 3

u + 3 = 0 : u = - 3

u + 3 = 0

Move

3

to the right side

u + 3 = 0

Subtract

3

from both sides

u + 3 - 3 = 0 - 3

Simplify

u = - 3

u = - 3

u + 3 < 0 : u < - 3

u + 3 < 0

Move

3

to the right side

u + 3 < 0

Subtract

3

from both sides

u + 3 - 3 < 0 - 3

Simplify

u < - 3

u < - 3

u + 3 > 0 : u > - 3

u + 3 > 0

Move

3

to the right side

u + 3 > 0

Subtract

3

from both sides

u + 3 - 3 > 0 - 3

Simplify

u > - 3

u > - 3

Find the signs of

u - 3

u - 3 = 0 : u = 3

u - 3 = 0

Move

3

to the right side

u - 3 = 0

Add

3

to both sides

u - 3 + 3 = 0 + 3

Simplify

u = 3

u = 3

u - 3 < 0 : u < 3

u - 3 < 0

Move

3

to the right side

u - 3 < 0

Add

3

to both sides

u - 3 + 3 < 0 + 3

Simplify

u < 3

u < 3

u - 3 > 0 : u > 3

u - 3 > 0

Move

3

to the right side

u - 3 > 0

Add

3

to both sides

u - 3 + 3 > 0 + 3

Simplify

u > 3

u > 3

Find the signs of

u + 1

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

u + 1 < 0 : u < - 1

u + 1 < 0

Move

1

to the right side

u + 1 < 0

Subtract

1

from both sides

u + 1 - 1 < 0 - 1

Simplify

u < - 1

u < - 1

u + 1 > 0 : u > - 1

u + 1 > 0

Move

1

to the right side

u + 1 > 0

Subtract

1

from both sides

u + 1 - 1 > 0 - 1

Simplify

u > - 1

u > - 1

Find the signs of

u - 1

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

u - 1 < 0 : u < 1

u - 1 < 0

Move

1

to the right side

u - 1 < 0

Add

1

to both sides

u - 1 + 1 < 0 + 1

Simplify

u < 1

u < 1

u - 1 > 0 : u > 1

u - 1 > 0

Move

1

to the right side

u - 1 > 0

Add

1

to both sides

u - 1 + 1 > 0 + 1

Simplify

u > 1

u > 1

Find singularity points

Find the zeros of the denominator

- (u + 1) (u - 1) : u = - 1, u = 1

- (u + 1) (u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u + 1 = 0 or u - 1 = 0

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

The solutions to the quadratic equation are:

u = - 1, u = 1

Summarize in a table:

u u + 3 u - 3 u + 1 u - 1 \frac{- u ( u + 3 ) ( u - 3 )}{- ( u + 1 ) ( u - 1 )} u < - 3 - - - - - - u = - 3 - 0 - - - 0 - 3 < u < - 1 - + - - - + u = - 1 - + - 0 - Undefined - 1 < u < 0 - + - + - - u = 0 0 + - + - 0 0 < u < 1 + + - + - + u = 1 + + - + 0 Undefined 1 < u < 3 + + - + + - u = 3 + + 0 + + 0 u > 3 + + + + + +

Identify the intervals that satisfy the required condition:

> 0

- 3 < u < - 1 or 0 < u < 1 or u > 3

- 3 < u < - 1 or 0 < u < 1 or u > 3

- 3 < u < - 1 or 0 < u < 1 or u > 3

Substitute back

u = tan (x)

- 3 < tan (x) < - 1 or 0 < tan (x) < 1 or tan (x) > 3

- 3 < tan (x) < - 1 : \frac{2 π}{3} + πn < x < \frac{3 π}{4} + πn

- 3 < tan (x) < - 1

a < u < b

then

a < u and u < b

- 3 < tan (x) and tan (x) < - 1

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

- 3 < tan (x)

Switch sides

tan (x) > - 3

tan (x) > a

then

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

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

Simplify

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}{2} + πn

tan (x) < - 1 : - \frac{π}{2} + πn < x < - \frac{π}{4} + πn

tan (x) < - 1

tan (x) < a

then

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

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

Simplify

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}{2} + πn < x < - \frac{π}{4} + πn

Combine the intervals

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

Merge Overlapping Intervals

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

0 < tan (x) < 1 : πn < x < \frac{π}{4} + πn

0 < tan (x) < 1

a < u < b

then

a < u and u < b

0 < tan (x) and tan (x) < 1

0 < tan (x) : πn < x < \frac{π}{2} + πn

0 < tan (x)

Switch sides

tan (x) > 0

tan (x) > a

then

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

arctan (0) + πn < x < \frac{π}{2} + πn

Simplify

arctan (0) : 0

arctan (0)

Use the following trivial identity:

arctan (0) = 0

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}

= 0

0 + πn < x < \frac{π}{2} + πn

Simplify

πn < x < \frac{π}{2} + πn

tan (x) < 1 : - \frac{π}{2} + πn < x < \frac{π}{4} + πn

tan (x) < 1

tan (x) < a

then

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

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

Simplify

arctan (1) : \frac{π}{4}

arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

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}{2} + πn < x < \frac{π}{4} + πn

Combine the intervals

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

Merge Overlapping Intervals

πn < x < \frac{π}{4} + πn

tan (x) > 3 : \frac{π}{3} + πn < x < \frac{π}{2} + πn

tan (x) > 3

tan (x) > a

then

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

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

Simplify

arctan (3) : \frac{π}{3}

arctan (3)

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}

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

Combine the intervals

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

Merge Overlapping Intervals

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

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