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

The general solution for tan^3(x)+tan^2(x)-3tan(x)-3=0 is x=(3pi)/4+pin,x=(2pi)/3+pin,x= pi/3+pin

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Popular >

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

Solution

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

Solution

x = \frac{3 π}{4} + πn, x = \frac{2 π}{3} + πn, x = \frac{π}{3} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n, x = 6 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

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

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

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

Factor

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

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

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

Factor out

- 3

from

- 3 u - 3 : - 3 (u + 1)

- 3 u - 3

Factor out common term

- 3

= - 3 (u + 1)

Factor out

u^{2}

from

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

u^{3} + u^{2}

Apply exponent rule:

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

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

= u u^{2} + u^{2}

Factor out common term

u^{2}

= u^{2} (u + 1)

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

Factor out common term

u + 1

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

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 = - 1, u = - 3, u = 3

Substitute back

u = tan (x)

tan (x) = - 1, tan (x) = - 3, tan (x) = 3

tan (x) = - 1, tan (x) = - 3, tan (x) = 3

tan (x) = - 1 : x = \frac{3 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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

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 = \frac{3 π}{4} + πn, x = \frac{2 π}{3} + πn, x = \frac{π}{3} + πn

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cos (\frac{θ}{2}) = \frac{1 + \frac{3}{5}}{2}

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cos (θ) = \frac{35}{37}

Frequently Asked Questions (FAQ)

What is the general solution for tan^3(x)+tan^2(x)-3tan(x)-3=0 ?
The general solution for tan^3(x)+tan^2(x)-3tan(x)-3=0 is x=(3pi)/4+pin,x=(2pi)/3+pin,x= pi/3+pin