What is the general solution for 2tan^3(x)-6tan(x)=0,0<= x<= pi ?

Q: What is the general solution for 2tan^3(x)-6tan(x)=0,0<= x<= pi ?

The general solution for 2tan^3(x)-6tan(x)=0,0<= x<= pi is x=0,x=pi,x=(2pi)/3 ,x= pi/3

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2tan^3(x)-6tan(x)=0,0<= x<= pi

Solution

2 tan^{3} (x) - 6 tan (x) = 0, 0 \leq x \leq π

Solution

x = 0, x = π, x = \frac{2 π}{3}, x = \frac{π}{3}

Degrees

x = 0^{\circ}, x = 18 0^{\circ}, x = 12 0^{\circ}, x = 6 0^{\circ}

Solution steps

2 tan^{3} (x) - 6 tan (x) = 0, 0 \leq x \leq π

Solve by substitution

2 tan^{3} (x) - 6 tan (x) = 0

Let:

tan (x) = u

2 u^{3} - 6 u = 0

2 u^{3} - 6 u = 0 : u = 0, u = - 3, u = 3

2 u^{3} - 6 u = 0

Factor

2 u^{3} - 6 u : 2 u (u + 3) (u - 3)

2 u^{3} - 6 u

Factor out common term

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

2 u^{3} - 6 u

Apply exponent rule:

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

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

= 2 u^{2} u - 6 u

Rewrite

6

2 \cdot 3

= 2 u^{2} u - 2 \cdot 3 u

Factor out common term

2 u

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

= 2 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)

= 2 u (u + 3) (u - 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

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 = 0, u = - 3, u = 3

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = - 3, tan (x) = 3

tan (x) = 0, tan (x) = - 3, tan (x) = 3

tan (x) = 0, 0 \leq x \leq π : x = 0, x = π

tan (x) = 0, 0 \leq x \leq π

General solutions for

tan (x) = 0

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 = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

Solutions for the range

0 \leq x \leq π

x = 0, x = π

tan (x) = - 3, 0 \leq x \leq π : x = \frac{2 π}{3}

tan (x) = - 3, 0 \leq x \leq π

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

0 \leq x \leq π

x = \frac{2 π}{3}

tan (x) = 3, 0 \leq x \leq π : x = \frac{π}{3}

tan (x) = 3, 0 \leq x \leq π

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

0 \leq x \leq π

x = \frac{π}{3}

Combine all the solutions

x = 0, x = π, x = \frac{2 π}{3}, x = \frac{π}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan^3(x)-6tan(x)=0,0<= x<= pi ?
The general solution for 2tan^3(x)-6tan(x)=0,0<= x<= pi is x=0,x=pi,x=(2pi)/3 ,x= pi/3