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

The general solution for tan^3(x)=tan(x) is x=pin,x=(3pi)/4+pin,x= pi/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^3(x)=tan(x)

Solution

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

Solution

x = πn, x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

u^{3} = u

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

u^{3} = u

Move

u

to the left side

u^{3} = u

Subtract

u

from both sides

u^{3} - u = u - u

Simplify

u^{3} - u = 0

u^{3} - u = 0

Factor

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

u^{3} - u

Factor out common term

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

u^{3} - u

Apply exponent rule:

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

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

= u^{2} u - u

Factor out common term

u

= u (u^{2} - 1)

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 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 are

u = 0, u = - 1, u = 1

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = - 1, tan (x) = 1

tan (x) = 0, tan (x) = - 1, tan (x) = 1

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) = - 1 : x = \frac{3 π}{4} + πn

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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

tan (x) = 1 : x = \frac{π}{4} + πn

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

Combine all the solutions

x = πn, x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Popular Examples

sqrt(3)-2sin(x)=0

3 - 2 sin (x) = 0

-9cos^2(θ)+14sin(θ)+14=0

- 9 cos^{2} (θ) + 14 sin (θ) + 14 = 0

cos(x)tan(x)=-7tan(x)

cos (x) tan (x) = - 7 tan (x)

12sin^2(θ)-9=0

12 sin^{2} (θ) - 9 = 0

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

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

Frequently Asked Questions (FAQ)

What is the general solution for tan^3(x)=tan(x) ?
The general solution for tan^3(x)=tan(x) is x=pin,x=(3pi)/4+pin,x= pi/4+pin