What is the general solution for tan^3(x)= 1/3 tan(x) ?

The general solution for tan^3(x)= 1/3 tan(x) is x=pin,x=-0.52359…+pin,x=0.52359…+pin

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tan^3(x)= 1/3 tan(x)

Solution

tan^{3} (x) = \frac{1}{3} tan (x)

Solution

x = πn, x = - 0.52359 \dots + πn, x = 0.52359 \dots + πn

Degrees

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

Solution steps

tan^{3} (x) = \frac{1}{3} tan (x)

Solve by substitution

tan^{3} (x) = \frac{1}{3} tan (x)

Let:

tan (x) = u

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

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

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

Move

\frac{1}{3} u

to the left side

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

Subtract

\frac{1}{3} u

from both sides

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

Simplify

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

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

Factor

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

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

Factor out common term

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

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

Apply exponent rule:

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

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

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

Factor out common term

u

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

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

Factor

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

u^{2} - \frac{1}{3}

Apply radical rule:

a = (a)^{2}

\frac{1}{3} = (\frac{1}{3})^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (u + \frac{1}{3}) (u - \frac{1}{3})

= u (u + \frac{1}{3}) (u - \frac{1}{3})

u (u + \frac{1}{3}) (u - \frac{1}{3}) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

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

Move

\frac{1}{3}

to the right side

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

Subtract

\frac{1}{3}

from both sides

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

Simplify

u = - \frac{1}{3}

u = - \frac{1}{3}

Solve

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

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

Move

\frac{1}{3}

to the right side

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

Add

\frac{1}{3}

to both sides

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

Simplify

u = \frac{1}{3}

u = \frac{1}{3}

The solutions are

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

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = - \frac{1}{3}, tan (x) = \frac{1}{3}

tan (x) = 0, tan (x) = - \frac{1}{3}, tan (x) = \frac{1}{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{1}{3} : x = arctan (- \frac{1}{3}) + πn

tan (x) = - \frac{1}{3}

Apply trig inverse properties

tan (x) = - \frac{1}{3}

General solutions for

tan (x) = - \frac{1}{3}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

tan (x) = \frac{1}{3}

Apply trig inverse properties

tan (x) = \frac{1}{3}

General solutions for

tan (x) = \frac{1}{3}

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

Combine all the solutions

x = πn, x = arctan (- \frac{1}{3}) + πn, x = arctan (\frac{1}{3}) + πn

Show solutions in decimal form

x = πn, x = - 0.52359 \dots + πn, x = 0.52359 \dots + πn

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3csc^2(x)+1.5cot(x)=15

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Frequently Asked Questions (FAQ)

What is the general solution for tan^3(x)= 1/3 tan(x) ?
The general solution for tan^3(x)= 1/3 tan(x) is x=pin,x=-0.52359…+pin,x=0.52359…+pin