What is the general solution for 2tan^3(x)-2/3 tan(x)=0 ?

The general solution for 2tan^3(x)-2/3 tan(x)=0 is x=pin,x=(5pi)/6+pin,x= pi/6+pin

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2tan^3(x)-2/3 tan(x)=0

Solution

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

Solution

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

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

2 u^{3} - \frac{2}{3} u = 0

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

2 u^{3} - \frac{2}{3} u = 0

Factor

2 u^{3} - \frac{2}{3} u : u (2 u + \frac{2}{3}) (2 u - \frac{2}{3})

2 u^{3} - \frac{2}{3} u

Factor out common term

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

2 u^{3} - \frac{2 u}{3}

Apply exponent rule:

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

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

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

Factor out common term

u

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

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

Factor

2 u^{2} - \frac{2}{3} : (2 u + \frac{2}{3}) (2 u - \frac{2}{3})

2 u^{2} - \frac{2}{3}

Rewrite

2 u^{2} - \frac{2}{3}

(2 u)^{2} - (\frac{2}{3})^{2}

2 u^{2} - \frac{2}{3}

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

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

Apply radical rule:

a = (a)^{2}

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

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

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

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

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

= (2 u + \frac{2}{3}) (2 u - \frac{2}{3})

= u (2 u + \frac{2}{3}) (2 u - \frac{2}{3})

u (2 u + \frac{2}{3}) (2 u - \frac{2}{3}) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 2 u + \frac{2}{3} = 0 or 2 u - \frac{2}{3} = 0

Solve

2 u + \frac{2}{3} = 0 : u = - \frac{3}{3}

2 u + \frac{2}{3} = 0

Move

\frac{2}{3}

to the right side

2 u + \frac{2}{3} = 0

Subtract

\frac{2}{3}

from both sides

2 u + \frac{2}{3} - \frac{2}{3} = 0 - \frac{2}{3}

Simplify

2 u = - \frac{2}{3}

2 u = - \frac{2}{3}

Divide both sides by

2

2 u = - \frac{2}{3}

Divide both sides by

2

\frac{2 u}{2} = \frac{- \frac{2}{3}}{2}

Simplify

\frac{2 u}{2} = \frac{- \frac{2}{3}}{2}

Simplify

\frac{2 u}{2} : u

\frac{2 u}{2}

Cancel the common factor:

2

= u

Simplify

\frac{- \frac{2}{3}}{2} : - \frac{3}{3}

\frac{- \frac{2}{3}}{2}

Apply the fraction rule:

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

= - \frac{\frac{2}{3}}{2}

\frac{2}{3} = \frac{2}{3}

\frac{2}{3}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2}{3}

= - \frac{\frac{2}{3}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= - \frac{2}{3 2}

Cancel the common factor:

2

= - \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

2 u - \frac{2}{3} = 0 : u = \frac{3}{3}

2 u - \frac{2}{3} = 0

Move

\frac{2}{3}

to the right side

2 u - \frac{2}{3} = 0

Add

\frac{2}{3}

to both sides

2 u - \frac{2}{3} + \frac{2}{3} = 0 + \frac{2}{3}

Simplify

2 u = \frac{2}{3}

2 u = \frac{2}{3}

Divide both sides by

2

2 u = \frac{2}{3}

Divide both sides by

2

\frac{2 u}{2} = \frac{\frac{2}{3}}{2}

Simplify

\frac{2 u}{2} = \frac{\frac{2}{3}}{2}

Simplify

\frac{2 u}{2} : u

\frac{2 u}{2}

Cancel the common factor:

2

= u

Simplify

\frac{\frac{2}{3}}{2} : \frac{3}{3}

\frac{\frac{2}{3}}{2}

\frac{2}{3} = \frac{2}{3}

\frac{2}{3}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2}{3}

= \frac{\frac{2}{3}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{2}{3 2}

Cancel the common factor:

2

= \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}

The solutions are

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

Substitute back

u = tan (x)

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

tan (x) = 0, tan (x) = - \frac{3}{3}, tan (x) = \frac{3}{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) = - \frac{3}{3} : x = \frac{5 π}{6} + πn

tan (x) = - \frac{3}{3}

General solutions for

tan (x) = - \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}

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

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

tan (x) = \frac{3}{3}

General solutions for

tan (x) = \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}

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan^3(x)-2/3 tan(x)=0 ?
The general solution for 2tan^3(x)-2/3 tan(x)=0 is x=pin,x=(5pi)/6+pin,x= pi/6+pin