What is the general solution for tan^3(x)-2tan(x)=tan(x) ?

The general solution for tan^3(x)-2tan(x)=tan(x) is x=pin,x=(2pi)/3+pin,x= pi/3+pin

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tan^3(x)-2tan(x)=tan(x)

Solution

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

Solution

x = πn, x = \frac{2 π}{3} + πn, x = \frac{π}{3} + πn

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

u^{3} - 2 u = u

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

u^{3} - 2 u = u

Move

u

to the left side

u^{3} - 2 u = u

Subtract

u

from both sides

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

Simplify

u^{3} - 3 u = 0

u^{3} - 3 u = 0

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)

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 : x = πn

tan (x) = 0

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

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

tan (x) = - 3

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

tan (x) = 3 : x = \frac{π}{3} + πn

tan (x) = 3

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

Combine all the solutions

x = πn, x = \frac{2 π}{3} + πn, x = \frac{π}{3} + πn

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Frequently Asked Questions (FAQ)

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