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4cos^4(x)-2cos^3(x)-4cos^2(x)+cos(x)+1=0

Solution

4 cos^{4} (x) - 2 cos^{3} (x) - 4 cos^{2} (x) + cos (x) + 1 = 0

Solution

x = 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{4} (x) - 2 cos^{3} (x) - 4 cos^{2} (x) + cos (x) + 1 = 0

Solve by substitution

4 cos^{4} (x) - 2 cos^{3} (x) - 4 cos^{2} (x) + cos (x) + 1 = 0

Let:

cos (x) = u

4 u^{4} - 2 u^{3} - 4 u^{2} + u + 1 = 0

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

4 u^{4} - 2 u^{3} - 4 u^{2} + u + 1 = 0

Factor

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

4 u^{4} - 2 u^{3} - 4 u^{2} + u + 1

Use the rational root theorem

a_{0} = 1, a_{n} = 4

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2, 4

Therefore, check the following rational numbers:

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

\frac{1}{1}

is a root of the expression, so factor out

u - 1

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

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

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

Divide

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

Divide the leading coefficients of the numerator

4 u^{4} - 2 u^{3} - 4 u^{2} + u + 1

and the divisor

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

Quotient = 4 u^{3}

Multiply

u - 1

4 u^{3} : 4 u^{4} - 4 u^{3}

Subtract

4 u^{4} - 4 u^{3}

from

4 u^{4} - 2 u^{3} - 4 u^{2} + u + 1

to get new remainder

Remainder = 2 u^{3} - 4 u^{2} + u + 1

Therefore

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

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

Divide

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

Divide the leading coefficients of the numerator

2 u^{3} - 4 u^{2} + u + 1

and the divisor

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

Quotient = 2 u^{2}

Multiply

u - 1

2 u^{2} : 2 u^{3} - 2 u^{2}

Subtract

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

from

2 u^{3} - 4 u^{2} + u + 1

to get new remainder

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

Therefore

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

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

Divide

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

Divide the leading coefficients of the numerator

- 2 u^{2} + u + 1

and the divisor

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

Quotient = - 2 u

Multiply

u - 1

- 2 u : - 2 u^{2} + 2 u

Subtract

- 2 u^{2} + 2 u

from

- 2 u^{2} + u + 1

to get new remainder

Remainder = - u + 1

Therefore

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

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

Divide

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

Divide the leading coefficients of the numerator

- u + 1

and the divisor

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

Quotient = - 1

Multiply

u - 1

- 1 : - u + 1

Subtract

- u + 1

from

- u + 1

to get new remainder

Remainder = 0

Therefore

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

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

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

Factor

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

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

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

Factor out

- 1

from

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

- 2 u - 1

Factor out common term

- 1

= - (2 u + 1)

Factor out

2 u^{2}

from

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

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

Apply exponent rule:

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

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

= 4 u u^{2} + 2 u^{2}

Rewrite

4

2 \cdot 2

= 2 \cdot 2 u u^{2} + 2 u^{2}

Factor out common term

2 u^{2}

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

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

Factor out common term

2 u + 1

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

Factor

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

2 u^{2} - 1

Rewrite

2 u^{2} - 1

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

2 u^{2} - 1

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

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

Rewrite

1

1^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

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

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

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

Simplify

\frac{2 u}{2} : u

\frac{2 u}{2}

Cancel the common factor:

2

= u

Simplify

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

\frac{- 1}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2}

Rationalize

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

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

Solve

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

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to 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

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

Simplify

\frac{2 u}{2} : u

\frac{2 u}{2}

Cancel the common factor:

2

= u

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

The solutions are

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

Substitute back

u = cos (x)

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

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

cos (x) = 1 : x = 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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 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{2}{2} : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

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

General solutions for

cos (x) = - \frac{2}{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{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

cos (x) = \frac{2}{2} : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

General solutions for

cos (x) = \frac{2}{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{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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