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

The general solution for 3tan^3(θ)=tan(θ) is θ=pin,θ=(5pi)/6+pin,θ= pi/6+pin

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Popular Trigonometry >

3tan^3(θ)=tan(θ)

Solution

3 tan^{3} (θ) = tan (θ)

Solution

θ = πn, θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n, θ = 15 0^{\circ} + 18 0^{\circ} n, θ = 3 0^{\circ} + 18 0^{\circ} n

Solution steps

3 tan^{3} (θ) = tan (θ)

Solve by substitution

3 tan^{3} (θ) = tan (θ)

Let:

tan (θ) = u

3 u^{3} = u

3 u^{3} = u : u = 0, u = - \frac{3}{3}, u = \frac{3}{3}

3 u^{3} = u

Move

u

to the left side

3 u^{3} = u

Subtract

u

from both sides

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

Simplify

3 u^{3} - u = 0

3 u^{3} - u = 0

Factor

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

3 u^{3} - u

Factor out common term

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

3 u^{3} - u

Apply exponent rule:

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

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

= 3 u^{2} u - u

Factor out common term

u

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

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

Factor

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

3 u^{2} - 1

Rewrite

3 u^{2} - 1

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

3 u^{2} - 1

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

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

Rewrite

1

1^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

3 u + 1 = 0 : u = - \frac{3}{3}

3 u + 1 = 0

Move

1

to the right side

3 u + 1 = 0

Subtract

1

from both sides

3 u + 1 - 1 = 0 - 1

Simplify

3 u = - 1

3 u = - 1

Divide both sides by

3

3 u = - 1

Divide both sides by

3

\frac{3 u}{3} = \frac{- 1}{3}

Simplify

\frac{3 u}{3} = \frac{- 1}{3}

Simplify

\frac{3 u}{3} : u

\frac{3 u}{3}

Cancel the common factor:

3

= u

Simplify

\frac{- 1}{3} : - \frac{3}{3}

\frac{- 1}{3}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{3}

Rationalize

- \frac{1}{3} : - \frac{3}{3}

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

u = - \frac{3}{3}

u = - \frac{3}{3}

u = - \frac{3}{3}

Solve

3 u - 1 = 0 : u = \frac{3}{3}

3 u - 1 = 0

Move

1

to the right side

3 u - 1 = 0

Add

1

to both sides

3 u - 1 + 1 = 0 + 1

Simplify

3 u = 1

3 u = 1

Divide both sides by

3

3 u = 1

Divide both sides by

3

\frac{3 u}{3} = \frac{1}{3}

Simplify

\frac{3 u}{3} = \frac{1}{3}

Simplify

\frac{3 u}{3} : u

\frac{3 u}{3}

Cancel the common factor:

3

= u

Simplify

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

u = \frac{3}{3}

u = \frac{3}{3}

u = \frac{3}{3}

The solutions are

u = 0, u = - \frac{3}{3}, u = \frac{3}{3}

Substitute back

u = tan (θ)

tan (θ) = 0, tan (θ) = - \frac{3}{3}, tan (θ) = \frac{3}{3}

tan (θ) = 0, tan (θ) = - \frac{3}{3}, tan (θ) = \frac{3}{3}

tan (θ) = 0 : θ = πn

tan (θ) = 0

General solutions for

tan (θ) = 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}

θ = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

tan (θ) = - \frac{3}{3} : θ = \frac{5 π}{6} + πn

tan (θ) = - \frac{3}{3}

General solutions for

tan (θ) = - \frac{3}{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}

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

tan (θ) = \frac{3}{3} : θ = \frac{π}{6} + πn

tan (θ) = \frac{3}{3}

General solutions for

tan (θ) = \frac{3}{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}

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

Combine all the solutions

θ = πn, θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

Graph

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What is the general solution for 3tan^3(θ)=tan(θ) ?
The general solution for 3tan^3(θ)=tan(θ) is θ=pin,θ=(5pi)/6+pin,θ= pi/6+pin