What is the general solution for tan^4(x)=tan(x) ?

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

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Popular >

tan^4(x)=tan(x)

Solution

tan^{4} (x) = tan (x)

Solution

x = πn, x = \frac{π}{4} + πn

Degrees

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

Solution steps

tan^{4} (x) = tan (x)

Solve by substitution

tan^{4} (x) = tan (x)

Let:

tan (x) = u

u^{4} = u

u^{4} = u : u = 0, u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u^{4} = u

Move

u

to the left side

u^{4} = u

Subtract

u

from both sides

u^{4} - u = u - u

Simplify

u^{4} - u = 0

u^{4} - u = 0

Factor

u^{4} - u : u (u^{3} - 1)

u^{4} - u

Apply exponent rule:

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

u^{4} = u^{3} u

= u^{3} u - u

Factor out common term

u

= u (u^{3} - 1)

u (u^{3} - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u^{3} - 1 = 0

Solve

u^{3} - 1 = 0 : u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u^{3} - 1 = 0

Move

1

to the right side

u^{3} - 1 = 0

Add

1

to both sides

u^{3} - 1 + 1 = 0 + 1

Simplify

u^{3} = 1

u^{3} = 1

For

x^{3} = f (a)

the solutions are

u = 1, u = \frac{- 1 + 3 i}{2}, u = \frac{- 1 - 3 i}{2}

Simplify

\frac{- 1 + 3 i}{2} : - \frac{1}{2} + i \frac{3}{2}

\frac{- 1 + 3 i}{2}

Rewrite

\frac{- 1 + 3 i}{2}

in standard complex form:

- \frac{1}{2} + \frac{3}{2} i

\frac{- 1 + 3 i}{2}

Apply the fraction rule:

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

\frac{- 1 + 3 i}{2} = - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3}{2} i

Simplify

\frac{- 1 - 3 i}{2} : - \frac{1}{2} - i \frac{3}{2}

\frac{- 1 - 3 i}{2}

Rewrite

\frac{- 1 - 3 i}{2}

in standard complex form:

- \frac{1}{2} - \frac{3}{2} i

\frac{- 1 - 3 i}{2}

Apply the fraction rule:

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

\frac{- 1 - 3 i}{2} = - \frac{1}{2} - \frac{3 i}{2}

= - \frac{1}{2} - \frac{3 i}{2}

= - \frac{1}{2} - \frac{3}{2} i

u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

The solutions are

u = 0, u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = 1, tan (x) = - \frac{1}{2} + i \frac{3}{2}, tan (x) = - \frac{1}{2} - i \frac{3}{2}

tan (x) = 0, tan (x) = 1, tan (x) = - \frac{1}{2} + i \frac{3}{2}, tan (x) = - \frac{1}{2} - i \frac{3}{2}

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{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

tan (x) = - \frac{1}{2} + i \frac{3}{2} :

No Solution

tan (x) = - \frac{1}{2} + i \frac{3}{2}

No Solution

tan (x) = - \frac{1}{2} - i \frac{3}{2} :

No Solution

tan (x) = - \frac{1}{2} - i \frac{3}{2}

No Solution

Combine all the solutions

x = πn, x = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

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