What is the general solution for cos^3(θ)+cos^2(θ)-cos(θ)-1=0 ?

The general solution for cos^3(θ)+cos^2(θ)-cos(θ)-1=0 is θ=pi+2pin,θ=2pin

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cos^3(θ)+cos^2(θ)-cos(θ)-1=0

Solution

cos^{3} (θ) + cos^{2} (θ) - cos (θ) - 1 = 0

Solution

θ = π + 2 πn, θ = 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{3} (θ) + cos^{2} (θ) - cos (θ) - 1 = 0

Solve by substitution

cos^{3} (θ) + cos^{2} (θ) - cos (θ) - 1 = 0

Let:

cos (θ) = u

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

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

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

Factor

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

u^{3} + u^{2} - u - 1

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

Factor out

- 1

from

- u - 1 : - (u + 1)

- u - 1

Factor out common term

- 1

= - (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)

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

Factor out common term

u + 1

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

Factor

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

u^{2} - 1

Rewrite

1

1^{2}

= u^{2} - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (u + 1) (u - 1)

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

Refine

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

(u + 1)^{2} (u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u + 1 = 0 or u - 1 = 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 - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

The solutions are

u = - 1, u = 1

Substitute back

u = cos (θ)

cos (θ) = - 1, cos (θ) = 1

cos (θ) = - 1, cos (θ) = 1

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

General solutions for

cos (θ) = - 1

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = π + 2 πn

θ = π + 2 πn

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

General solutions for

cos (θ) = 1

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Combine all the solutions

θ = π + 2 πn, θ = 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^3(θ)+cos^2(θ)-cos(θ)-1=0 ?
The general solution for cos^3(θ)+cos^2(θ)-cos(θ)-1=0 is θ=pi+2pin,θ=2pin