What is the general solution for (tan(x))^5-9tan(x)=0 ?

The general solution for (tan(x))^5-9tan(x)=0 is x=pin,x=(2pi)/3+pin,x= pi/3+pin

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(tan(x))^5-9tan(x)=0

Solution

(tan (x))^{5} - 9 tan (x) = 0

Solution

x = πn, x = \frac{2 π}{3} + πn, x = \frac{π}{3} + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n, x = 6 0^{\circ} + 18 0^{\circ} n

Solution steps

(tan (x))^{5} - 9 tan (x) = 0

Solve by substitution

(tan (x))^{5} - 9 tan (x) = 0

Let:

tan (x) = u

u^{5} - 9 u = 0

u^{5} - 9 u = 0 : u = 0, u = 3 i, u = - 3 i, u = - 3, u = 3

u^{5} - 9 u = 0

Factor

u^{5} - 9 u : u (u^{2} + 3) (u + 3) (u - 3)

u^{5} - 9 u

Factor out common term

u : u (u^{4} - 9)

u^{5} - 9 u

Apply exponent rule:

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

u^{5} = u^{4} u

= u^{4} u - 9 u

Factor out common term

u

= u (u^{4} - 9)

= u (u^{4} - 9)

Factor

u^{4} - 9 : (u^{2} + 3) (u + 3) (u - 3)

u^{4} - 9

Rewrite

u^{4} - 9

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

u^{4} - 9

Rewrite

9

3^{2}

= u^{4} - 3^{2}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

u^{4} = (u^{2})^{2}

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

u^{2} - 3

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (u + 3) (u - 3)

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

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

u (u^{2} + 3) (u + 3) (u - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u^{2} + 3 = 0 or u + 3 = 0 or u - 3 = 0

Solve

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

u^{2} + 3 = 0

Move

3

to the right side

u^{2} + 3 = 0

Subtract

3

from both sides

u^{2} + 3 - 3 = 0 - 3

Simplify

u^{2} = - 3

u^{2} = - 3

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 3, u = - - 3

Simplify

- 3 : 3 i

- 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

Simplify

- - 3 : - 3 i

- - 3

Simplify

- 3 : 3 i

- 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

= - 3 i

u = 3 i, u = - 3 i

Solve

u + 3 = 0 : u = - 3

u + 3 = 0

Move

3

to the right side

u + 3 = 0

Subtract

3

from both sides

u + 3 - 3 = 0 - 3

Simplify

u = - 3

u = - 3

Solve

u - 3 = 0 : u = 3

u - 3 = 0

Move

3

to the right side

u - 3 = 0

Add

3

to both sides

u - 3 + 3 = 0 + 3

Simplify

u = 3

u = 3

The solutions are

u = 0, u = 3 i, u = - 3 i, u = - 3, u = 3

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = 3 i, tan (x) = - 3 i, tan (x) = - 3, tan (x) = 3

tan (x) = 0, tan (x) = 3 i, tan (x) = - 3 i, tan (x) = - 3, tan (x) = 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) = 3 i :

No Solution

tan (x) = 3 i

No Solution

tan (x) = - 3 i :

No Solution

tan (x) = - 3 i

No Solution

tan (x) = - 3 : x = \frac{2 π}{3} + πn

tan (x) = - 3

General solutions for

tan (x) = - 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{2 π}{3} + πn

x = \frac{2 π}{3} + πn

tan (x) = 3 : x = \frac{π}{3} + πn

tan (x) = 3

General solutions for

tan (x) = 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}{3} + πn

x = \frac{π}{3} + πn

Combine all the solutions

x = πn, x = \frac{2 π}{3} + πn, x = \frac{π}{3} + πn

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sin(2θ)=cos(θ),0<= θ<2pi

sin (2 θ) = cos (θ), 0 \leq θ < 2 π

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Frequently Asked Questions (FAQ)

What is the general solution for (tan(x))^5-9tan(x)=0 ?
The general solution for (tan(x))^5-9tan(x)=0 is x=pin,x=(2pi)/3+pin,x= pi/3+pin