What is the general solution for 4cos^3(x)-7cos(x)-3=0 ?

The general solution for 4cos^3(x)-7cos(x)-3=0 is x=pi+2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin

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4cos^3(x)-7cos(x)-3=0

Solution

4 cos^{3} (x) - 7 cos (x) - 3 = 0

Solution

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

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{3} (x) - 7 cos (x) - 3 = 0

Solve by substitution

4 cos^{3} (x) - 7 cos (x) - 3 = 0

Let:

cos (x) = u

4 u^{3} - 7 u - 3 = 0

4 u^{3} - 7 u - 3 = 0 : u = - 1, u = - \frac{1}{2}, u = \frac{3}{2}

4 u^{3} - 7 u - 3 = 0

Factor

4 u^{3} - 7 u - 3 : (u + 1) (2 u + 1) (2 u - 3)

4 u^{3} - 7 u - 3

Use the rational root theorem

a_{0} = 3, a_{n} = 4

The dividers of

a_{0} : 1, 3,

The dividers of

a_{n} : 1, 2, 4

Therefore, check the following rational numbers:

\pm \frac{1 , 3}{1 , 2 , 4}

- \frac{1}{1}

is a root of the expression, so factor out

u + 1

= (u + 1) \frac{4 u ^{3} - 7 u - 3}{u + 1}

\frac{4 u ^{3} - 7 u - 3}{u + 1} = 4 u^{2} - 4 u - 3

\frac{4 u ^{3} - 7 u - 3}{u + 1}

Divide

\frac{4 u ^{3} - 7 u - 3}{u + 1} : \frac{4 u ^{3} - 7 u - 3}{u + 1} = 4 u^{2} + \frac{- 4 u ^{2} - 7 u - 3}{u + 1}

Divide the leading coefficients of the numerator

4 u^{3} - 7 u - 3

and the divisor

u + 1 : \frac{4 u ^{3}}{u} = 4 u^{2}

Quotient = 4 u^{2}

Multiply

u + 1

4 u^{2} : 4 u^{3} + 4 u^{2}

Subtract

4 u^{3} + 4 u^{2}

from

4 u^{3} - 7 u - 3

to get new remainder

Remainder = - 4 u^{2} - 7 u - 3

Therefore

\frac{4 u ^{3} - 7 u - 3}{u + 1} = 4 u^{2} + \frac{- 4 u ^{2} - 7 u - 3}{u + 1}

= 4 u^{2} + \frac{- 4 u ^{2} - 7 u - 3}{u + 1}

Divide

\frac{- 4 u ^{2} - 7 u - 3}{u + 1} : \frac{- 4 u ^{2} - 7 u - 3}{u + 1} = - 4 u + \frac{- 3 u - 3}{u + 1}

Divide the leading coefficients of the numerator

- 4 u^{2} - 7 u - 3

and the divisor

u + 1 : \frac{- 4 u ^{2}}{u} = - 4 u

Quotient = - 4 u

Multiply

u + 1

- 4 u : - 4 u^{2} - 4 u

Subtract

- 4 u^{2} - 4 u

from

- 4 u^{2} - 7 u - 3

to get new remainder

Remainder = - 3 u - 3

Therefore

\frac{- 4 u ^{2} - 7 u - 3}{u + 1} = - 4 u + \frac{- 3 u - 3}{u + 1}

= 4 u^{2} - 4 u + \frac{- 3 u - 3}{u + 1}

Divide

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

Divide the leading coefficients of the numerator

- 3 u - 3

and the divisor

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

Quotient = - 3

Multiply

u + 1

- 3 : - 3 u - 3

Subtract

- 3 u - 3

from

- 3 u - 3

to get new remainder

Remainder = 0

Therefore

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

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

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

Factor

4 u^{2} - 4 u - 3 : (2 u + 1) (2 u - 3)

4 u^{2} - 4 u - 3

Break the expression into groups

4 u^{2} - 4 u - 3

Definition

Factors of

12 : 1, 2, 3, 4, 6, 12

12

Divisors (Factors)

Find the Prime factors of

12 : 2, 2, 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

Multiply the prime factors of

12 : 4, 6

2 \cdot 2 = 4

2 \cdot 3 = 6

4, 6

4, 6

Add the prime factors:

2, 3

Add 1 and the number

12

itself

1, 12

The factors of

12

1, 2, 3, 4, 6, 12

Negative factors of

12 : - 1, - 2, - 3, - 4, - 6, - 12

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 3, - 4, - 6, - 12

For every two factors such that

u * v = - 12,

check if

u + v = - 4

Check

u = 1, v = - 12 : u * v = - 12, u + v = - 11 \Rightarrow

FalseCheck

u = 2, v = - 6 : u * v = - 12, u + v = - 4 \Rightarrow

True

u = 2, v = - 6

Group into

(a x^{2} + ux) + (vx + c)

(4 u^{2} + 2 u) + (- 6 u - 3)

= (4 u^{2} + 2 u) + (- 6 u - 3)

Factor out

2 u

from

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

4 u^{2} + 2 u

Apply exponent rule:

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

u^{2} = uu

= 4 uu + 2 u

Rewrite

4

2 \cdot 2

= 2 \cdot 2 uu + 2 u

Factor out common term

2 u

= 2 u (2 u + 1)

Factor out

- 3

from

- 6 u - 3 : - 3 (2 u + 1)

- 6 u - 3

Rewrite

6

3 \cdot 2

= - 3 \cdot 2 u - 3

Factor out common term

- 3

= - 3 (2 u + 1)

= 2 u (2 u + 1) - 3 (2 u + 1)

Factor out common term

2 u + 1

= (2 u + 1) (2 u - 3)

= (u + 1) (2 u + 1) (2 u - 3)

(u + 1) (2 u + 1) (2 u - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u + 1 = 0 or 2 u + 1 = 0 or 2 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

2 u + 1 = 0 : u = - \frac{1}{2}

2 u + 1 = 0

Move

1

to the right side

2 u + 1 = 0

Subtract

1

from both sides

2 u + 1 - 1 = 0 - 1

Simplify

2 u = - 1

2 u = - 1

Divide both sides by

2

2 u = - 1

Divide both sides by

2

\frac{2 u}{2} = \frac{- 1}{2}

Simplify

u = - \frac{1}{2}

u = - \frac{1}{2}

Solve

2 u - 3 = 0 : u = \frac{3}{2}

2 u - 3 = 0

Move

3

to the right side

2 u - 3 = 0

Add

3

to both sides

2 u - 3 + 3 = 0 + 3

Simplify

2 u = 3

2 u = 3

Divide both sides by

2

2 u = 3

Divide both sides by

2

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

Simplify

u = \frac{3}{2}

u = \frac{3}{2}

The solutions are

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

Substitute back

u = cos (x)

cos (x) = - 1, cos (x) = - \frac{1}{2}, cos (x) = \frac{3}{2}

cos (x) = - 1, cos (x) = - \frac{1}{2}, cos (x) = \frac{3}{2}

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 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}

x = π + 2 πn

x = π + 2 πn

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = - \frac{1}{2}

General solutions for

cos (x) = - \frac{1}{2}

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}

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

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

cos (x) = \frac{3}{2} :

No Solution

cos (x) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^3(x)-7cos(x)-3=0 ?
The general solution for 4cos^3(x)-7cos(x)-3=0 is x=pi+2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin