What is the general solution for tan^3(x)-tan^2(x)-3tan(x)+3=0,tan(x)>1 ?

The general solution for tan^3(x)-tan^2(x)-3tan(x)+3=0,tan(x)>1 is No Solution for x\in\mathbb{R}

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Popular >

tan^3(x)-tan^2(x)-3tan(x)+3=0,tan(x)>1

Solution

tan^{3} (x) - tan^{2} (x) - 3 tan (x) + 3 = 0, tan (x) > 1

Solution

No Solution for x \in R

Solution steps

tan^{3} (x) - tan^{2} (x) - 3 tan (x) + 3 = 0, tan (x) > 1

Solve by substitution

tan^{3} (x) - tan^{2} (x) - 3 tan (x) + 3 = 0

Let:

tan (x) = u

u^{3} - u^{2} - 3 u + 3 = 0

u^{3} - u^{2} - 3 u + 3 = 0 : u = 1, u = - 3, u = 3

u^{3} - u^{2} - 3 u + 3 = 0

Factor

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

u^{3} - u^{2} - 3 u + 3

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

Factor out

- 3

from

- 3 u + 3 : - 3 (u - 1)

- 3 u + 3

Factor out common term

- 3

= - 3 (u - 1)

Factor out

u^{2}

from

u^{3} - u^{2} : u^{2} (u - 1)

u^{3} - u^{2}

Apply exponent rule:

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

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

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

Factor out common term

u^{2}

= u^{2} (u - 1)

= - 3 (u - 1) + u^{2} (u - 1)

Factor out common term

u - 1

= (u - 1) (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 - 1) (u + 3) (u - 3)

(u - 1) (u + 3) (u - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u - 1 = 0 or u + 3 = 0 or u - 3 = 0

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

Solve

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

Solve

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

The solutions are

u = 1, u = - 3, u = 3

Substitute back

u = tan (x)

tan (x) = 1, tan (x) = - 3, tan (x) = 3

tan (x) = 1, tan (x) = - 3, tan (x) = 3

tan (x) = 1, tan (x) > 1 :

No Solution

tan (x) = 1, tan (x) > 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solutions for the range

tan (x) > 1

No Solution

tan (x) = - 3, tan (x) > 1 :

No Solution

tan (x) = - 3, tan (x) > 1

General solutions for

tan (x) = - 3

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

Solutions for the range

tan (x) > 1

No Solution

tan (x) = 3, tan (x) > 1 :

No Solution

tan (x) = 3, tan (x) > 1

General solutions for

tan (x) = 3

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

Solutions for the range

tan (x) > 1

No Solution

Combine all the solutions

No Solution for x \in R

Popular Examples

sin(20-θ)-cos(10+3θ)=0

sin (2 0^{\circ} - θ) - cos (1 0^{\circ} + 3 θ) = 0

tan(x)= 1/1

tan (x) = \frac{1}{1}

cos(2θ)+3sin(θ)=2

cos (2 θ) + 3 sin (θ) = 2

0=2sin(2x)-sin(x)

0 = 2 sin (2 x) - sin (x)

sec^2(x/3)=2

sec^{2} (\frac{x}{3}) = 2

Frequently Asked Questions (FAQ)

What is the general solution for tan^3(x)-tan^2(x)-3tan(x)+3=0,tan(x)>1 ?
The general solution for tan^3(x)-tan^2(x)-3tan(x)+3=0,tan(x)>1 is No Solution for x\in\mathbb{R}